This is in some sense a rehash of Joel's answer and Todor's answer (and it's also repeating some of the content of a recent answer of mine to another question), but I think this presentation of the relevant fact is a bit more impactful.
Let's say I write down a big polynomial $p(\ell,x,y,z,w,t,n,k_1,\dots,k_m)$ with integer coefficients and then I define the following function: $$ f(\ell,x,y,z) = \inf_{w \in \mathbb{R}}\sup_{t \in \mathbb{R}}\inf_{n \in \mathbb{N}}\sup_{k_1,\dots,k_m \in \mathbb{N}} p(\ell,x,y,z,w,t,n,k_1,\dots,k_m). $$ You might not really see why I would be doing this, but this is certainly within the general realm of functions one might imagine seeing in real analysis.
I think the following statement is so reasonable-sounding you wouldn't even really think about it. (As Todor said in his answer, this really feels like it should be true, morally speaking.)
Reasonable-sounding statement. The functions $f(1,x,y,z),f(2,x,y,z),\dots,f(64,x,y,z)$ are not the indicator functions of the pieces of a Banach–Tarski paradoxical decomposition of the unit ball.
This is independent of $\mathsf{ZFC}$. In particular, there is a specific polynomial $p$ of the above form such that $V=L$ implies the negation of this reasonable-sounding statement, but it is also consistent with $\mathsf{ZFC}$ that all functions defined in the same manner as $f$ are (universally) measurable. (Moreover, this is flat out implied by things like projective determinacy or the existence of various large cardinals.)
The same notes as in my previous answer apply here. The polynomial has degree $7$. Using some not-entirely-published results of Jones on universal Diophantine equations, I can get $m$ in the above expression down to $76$, but without those I can only do $m \approx 5^{60}$.
As far as I can tell, the coefficients needed to do something like this have never actually been computed, but it should in principle be possible to do with a computer. They would be pretty big, but I think not outside the realm of something that could be written down in, say, a book. Moreover, using a higher degree universal polynomial, the polynomial $p$ (but not all of its coefficients) could actually fit into a MathOverflow answer (although it would also be degree $3 + 5^{60}$).
The number of pieces in the decomposition can probably be brought down to the known optimal number of $5$, but I've only checked the standard construction of the Banach–Tarski paradox carefully.