As noted in a comment-answer by te4, $\chi(H)=2$; a proper $2$-coloring is obtained by coloring the vertex $\varnothing$ (or the vertex $\omega$) red and all other vertices blue.
More generally, if the poset $(P,\le)$ has a maximal antichain consisting of a single element, and if $|P|\gt1$, then the hypergraph $(P,E)$, where $E$ is the set of all maximal chains in $(P,\le)$, has chromatic number $2$.
In view of this trivial answer, I suspect that the question was misstated. Perhaps you meant to ask about the maximal chains in the set of all nonempty proper subsets of $\omega$? In that case the chromatic number is still $2$ but the coloring is slightly less trivial. More generally:
Theorem. Suppose $h,k\lt\omega$. If $\{X\subseteq\omega:|X|=h\ \text{ or }\ |\omega\setminus X|=k\}\subseteq V\subseteq\mathcal P(\omega)$, and if $E$ is the set of all maximal chains in the poset $(V,\subseteq)$, then the hypergraph $H=(V,E)$ has chromatic number $\chi(H)=2$.
Proof. Plainly $\chi(H)\gt1$; we have to show that $H$ is $2$-colorable. Write $1=\sum_{n=1}^\infty a_n$ where $0\lt a_n\lt\frac1{2\max(h,k)}$ for all $n$. For $X\subseteq\omega$ let $m(X)=\sum_{n\in X}a_n$. I claim that every maximal chain $\mathcal C$ contains vertices $X$ and $Y$ with $m(X)\lt\frac12\lt m(Y)$.
Assume for a contradiction that $m(X)\ge\frac12$ for all $X\in\mathcal C$. Let $X_0=\bigcap\mathcal C$; then $m(X_0)\ge\frac12$, whence $|X_0|\gt h$. It follows that $X_0$ has a proper subset $X_0^\prime$ of cardinality $h$, and so $\mathcal C\cup\{X_0^\prime\}$ is a chain in $(V,\subseteq)$ properly extending $\mathcal C$, contradicting the maximality of $\mathcal C$.
Dually, assuming that $m(X)\le\frac12$ for all $X\in\mathcal C$, let $X_1=\bigcup\mathcal C$. Then $m(X_1)\le\frac12$, whence $|\omega\setminus X_1|\gt k$, etc.