Here by "antichain" I mean a set of elements that have pairwise-trivial meets, not merely ones that are pairwise-incomparable. Clearly, every atomless Boolean algebra has antichains in this sense of unbounded finite cardinality. However, every natural way I can think to prove that infinite antichains must always exist invokes some form of Choice.
I'd also be interested to know, if the answer is "yes," whether it is also yes when we only require atomlessness and not also completeness.
I'll give a characterization of the principle "every complete atomless algebra has an infinite antichain."
Conventions. We take "algebra" to mean an atomless Boolean algebra (and similarly for variations like "subalgebra"). An $\omega$-tree is a poset $(T, <)$ with a maximum for which each up-set $t \uparrow$ is a chain and $\{|t \uparrow |: t \in T\}=\omega.$ For $R \subset X \times Y,$ let $R_x$ and $R^y$ denote vertical and horizontal sections, respectively, and $\pi_i$ is the projection to coordinate $i.$
Adapting the terminology of Blackadar-Farah-Karagila, a set $A$ is Cohen infinite set if $[A]^{<\omega}$ is Dedekind infinite.
TFAE:
Proof. We let (n') denote the restriction of principle (n) to complete algebras, e.g. (2') is "every complete algebra is Dedekind infinite."
(2) $\, \leftrightarrow$ (3) $\, \leftrightarrow$ (4) $\, \rightarrow$ (1)
$\downarrow \: \qquad \downarrow \: \qquad \downarrow \: \qquad \downarrow$
(2') $\leftrightarrow$ (3') $\leftrightarrow$ (4') $\rightarrow$ (1')
This follows from verifying for a fixed algebra $B$ the implications "$B$ has a countably infinite antichain" $\leftrightarrow$ "$B$ has a countable infinite chain" $\leftrightarrow$ "$B$ is Dedekind infinite $\rightarrow$ "$B$ has an infinite antichain."
(6) $\leftrightarrow$ (4')
$\rightarrow$ is clear. For $\leftarrow,$ let $\langle x_n \rangle$ be an antichain in a complete algebra $B.$ The points of the form $y_{a,b}:=\bigvee_{k<\omega} x_{a+bk}$ generate a countable subalgebra.
(1') $\rightarrow$ (7)
If $A$ is an infinite Cohen finite set, a generalization of the sunflower argument in Andreas' answer shows that every point in $\mathrm{Add}(A, 1)$ has finite support, and its chains and antichains are finite.
Now we show (1') implies Form 216. Suppose $(T, <)$ is an $\omega$-tree of height $\omega$ with no infinite chain or antichain. Pruning $T$ yields a subtree $(T', <)$ of height $\omega$ which splits below every node. Let $B=\mathrm{RO}(T')$ i.e. the complete algebra which forces a generic branch $b$ through $T'.$
Suppose $X \subset B$ is an infinite antichain. For $x \in X,$ let $$x'=\{y \in T': \forall z \in T' (|y \uparrow^T| \le |z \uparrow^T|)\}.$$ Then $\{x': x \in X\}$ is an infinite antichain in $T,$ contradiction.
Suppose $C \subset B$ is an infinite chain, which we may assume to be closed under inf and sup of subsets. Define a map $f: T \rightarrow C$ by $f(y)=\sup \{x \in C: y <^B x\}.$ Then $C':=f"(T)$ is countable dense subset of $(C, <^B),$ so $C'$ is a countably infinite chain. Then $B$ has a countably infinite antichain, contradicting the above.
(7) $\rightarrow$ (2)
Assume (7). Let $B$ be an algebra. Let $\langle s_n : n<\omega \rangle$ be a sequence of disjoint nonempty finite subsets of $B \setminus \{0, 1\}.$ Let $t_n$ be the set of minimal nonzero Boolean combinations of points in $\bigcup_{k<n} s_k.$ Let $$T=\bigcup_{n<\omega} (\{n\} \times t_n) \subset \omega^* \times B$$ and equip $T$ with the lexicographical order. Then $T$ is an $\omega$-tree and $\langle T_n \rangle$ is a sequence of finite partitions of $B$ ordered by refinement.
Let $T'=\{(n, u) \in T: n=\min(T^u) \vee |T^u|<\aleph_0\}.$ Then $T'$ is an $\omega$-tree and thus has an infinite chain or antichain.
If $T'$ has an infinite chain $C,$ then $\pi_1"(C)$ is a countably infinite chain. Suppose $T'$ has an infinite antichain $X.$ Let $\langle k_n \rangle$ enumerate the $k$ such that $X \cap T'_k \neq \emptyset.$ Then $n \mapsto \bigvee (A \cap T'_{k_n})$ is an injection from $\omega$ into $B. \square$
Note that (2) shows that this principle follows from Form 9 (every infinite set is Dedekind infinite). However it is incomparable to a slightly weaker principle:
Theorem. Form 247 is incomparable to Form 10 (a countable union of finite sets is countable) over ZF.
Proof. In the Cohen model, there is an infinite Dedekind finite set of reals, which is necessarily Cohen finite. Thus form 247 fails while Form 10 holds in this model.
In Harry West's permutation model, Form 216 holds, Form 10 fails, and every infinite set is Cohen finite (immediate from Multiple Choice holding in this model, analogously to [1]). The assertions that Form 10 fails and a Dedekind finite set cannot support an atomless algebra are injectively bounded, so their conjunction transfers to ZF. $\square$
Theorem. The assertion that every algebra has a countable subalgebra is strictly stronger than Form 247. It follows from DC, but it does not follow from $\mathrm{AC}_{\omega} + \mathrm{BPI}.$
Proof. Howard/Rubin Model I ($\mathcal{N}38$ in Consequences of the Axiom of Choice, introduced in [3]) satisfies $\mathrm{ZFA} + \mathrm{AC}_{\omega} + \mathrm{BPI}$ and has a dense linear order $(A, <)$ with no descending sequence. Let $U$ be the set of finite unions of intervals in $A.$ Then $B=(U/[A]^{<\omega}, \subset)$ is an algebra.
Suppose $S$ is a countable subalgebra. Let $\langle e_n \rangle$ enumerate the end points of all nontrivial intervals $I$ such that for some $s \in S,$ $I$ is a maximal subinterval almost contained in $s.$ Then $(\langle e_n \rangle, <)$ is a countable well-order, and any $s \in S$ for which $\mathrm{ot}(\{e_n: [e_n, e_{n+1}] \subset^* s\})$ is minimized is an atom, contradiction.
These facts all transfer to ZF by Pincus' transfer theorem for BPI ([2]). $\square$
[1] Lévy, A., Axioms of multiple choice, Fundam. Math. 50, 475-483 (1962). ZBL0134.24805. Link
[2] Pincus, David, Adding dependent choice, Ann. Math. Logic 11, 105-145 (1977). ZBL0365.02052.
[3] Howard, P. and J. Rubin., The Boolean Prime Ideal Theorem Plus Countable Choice Do Not Imply Dependent Choice. Math. Log. Q. 42 (1996): 410-420.
[4] Karagila, A. and Schlicht, P. How to have more things by forgetting how to count them. Proc. R. Soc. A476: 20190782. arXiv link
I think that the answer is no.
I recently learned from a paper of Bodor, Braunfeld, and Hanson that the following is a theorem of Plotkin. For any $\omega$-categorical theory $T$ in a countable language there is a model $M$ of $\mathrm{ZF}$ such that there is a model $N$ of $T$ in $M$ such that the only subsets of $N^n$ in $M$ are those definable in $N$. So we apply this to the case when $N$ is the countable atomless boolean algebra, as this is an $\omega$-categorical structure.
Suppose that $B$ is the countable atomless boolean algebra. We just need to show that $B$ does not define an infinite antichain. (And we can use $\mathrm{ZFC}$.) I will just give a sketch. Suppose that $X$ is an infinite antichain definable over some finite set $A$ of parameters. Reduce to the case when $A$ is a partition. Let $S$ be the Stone space of $B$, so $S$ is just the Cantor set, $A$ is a partition of $S$ into clopen sets, and $X$ is an infinite family of pairwise-disjoint clopen subsets of $S$. Then some piece $P$ of the partition must intersect infinitely many elements of $X$. After intersection each member of $X$ with $P$ we may suppose that every member of $X$ is contained in $P$. Now $P$ is also homeomorphic to the Cantor set, so there is a homeomorphism $P \to P$ that does not preserve $X$, hence there is a homeomorphism $S \to S$ that preserves each element of $A$ but does not preserve $X$. By Stone duality there is an automorphism of $B$ that preserves every $a \in A$ but does not preserve $X$, contradiction.
We can avoid Stone duality by considering the localization of $B$ at $P$, defining an automorphism of this localization, and extending it to an autormorphism of $B$. This is probably less familiar.
Edit: Looks like Noah formulated essentially the same construction as I was writing this.
Finally, I think that this is the fifth reason I've seen to be interested in $\omega$-categorical structures, and more specifically in finitely homogeneous structures. They are interesting for combinatorial reasons because they are equivalent to Fraïssé classes, for group-theoretic reasons because their automorphism groups are exactly the ogliomorphic groups, for logical reasons because finitely homogeneous structures are exactly the countable structures with quantifier elimination in finite relational languages, for representation-theoretic reasons because they can be used to construct interesting tensor categories, and now for set-theoretic reasons because they are connected to models of $\mathrm{ZF}$ that violate choice.
In addition to Erik's nice abstract answer, here is a more concrete example: If there is an amorphous set (i.e. an infinite set with only finite and cofinite subsets) then there is a complete atomless Boolean algebra with no infinite antichains.
Let $A$ be an amorphous set and let $\mathbb P$ be the partial order of finite partial functions $p\colon A\rightarrow 2$ ordered by $\supseteq$ (this is the forcing that adds an infinite coinfinite subset to $A$). I claim that the Boolean completion $\mathbb B$ of $\mathbb P$ does not have infinite antichains.
First, let's see that $\mathbb P$ does not have infinite antichains. Suppose $\mathcal A$ is a counterexample.
Claim: For $n<\omega$, $\mathcal A_n:= \{p\in\mathcal A\mid\vert\mathrm{dom}(p)\vert=n\}$ is finite.
Proof: If not, we can apply the (finite) sunflower lemma to $\mathcal D=\{\mathrm{dom}(p)\mid p\in\mathcal A_n\}$ and find a sunflower $\mathcal S$ contained in $\mathcal D$ with $2^{n-1}+1$ many petals. For each $d\in\mathcal S$ choose $p_d\in\mathcal A_n$ with $\mathrm{dom}(p_d)=d$. Then $\{p_d\mid d\in\mathcal S\}$ is clearly not an antichain, contradiction. $\Box$
So $D_n:=\bigcup \{\mathrm{dom}(p)\mid p\in \mathcal A_n\}$ is finite. As $\mathcal A$ is infinite, $\bigcup_{n<\omega} D_n$ must be infinite. From this, we can easily construct a sequence $\langle a_n\mid n<\omega\rangle$ of finite nonempty pairwise disjoint subsets of $A$. But then $\bigcup_{n<\omega}a_{2n}$ is an infinite coinfinite subset of $A$, contradiction.
If we lived in a $\mathrm{ZFC}$ world, the completion $\mathbb B$ could not contain larger antichains than $\mathbb P$ itself and we would be done. However, its not clear to me whether this is true in $\mathrm{ZF}$ only. So let's see that $\mathbb B$ does not have infinite antichains either.
There is a function $f\colon \mathbb B\rightarrow [\mathbb P]^{<\omega}\setminus\{\emptyset\}$ so that $p\leq b$ for all $p\in f(b)$ and $b\in\mathbb B$. For example $f(b)=\{p\in\mathbb P\mid p\leq b\wedge \vert\mathrm{dom}(p)\vert=k\}$ works where $k$ is least such there is some $q\in\mathbb P$ with $q\leq b$ and $\vert\mathrm{dom}(q)\vert=k$.
Using $f$, we can replace an antichain $\mathcal A$ in $\mathbb B$ by a collection $\mathcal A'\subseteq [\mathbb P]^{<\omega}\setminus\{\emptyset\}$ of the same size with the property that for any $a\neq b$ in $\mathcal A'$, $p\perp q$ holds for all $p\in a$ and $q\in b$. The argument which shows that $\mathbb P$ does not have infinite antichains is good enough to prove that such a set $\mathcal A'$ must be finite.
