Given $\mathcal{A}\subseteq\omega^\omega$, the Banach-Mazur game with payoff set $\mathcal{A}$ consists of players $1$ and $2$ alternately playing nonempty finite strings of naturals with player $1$ winning iff the resulting infinite concatenation is in $\mathcal{A}$. For any bounded level of the projective hierarchy we can formalize this notion in the language of second-order arithmetic to define the sentence $\Sigma^1_n$-$\mathsf{BM}$ for each specific $n\in\omega$. (Here, $\Sigma^1_n$ is used in the boldface sense - that is, set parameters are allowed.)
It is well-known that even $\Sigma^1_2$-$\mathsf{BM}$ is not provable in $\mathsf{ZFC}$. On the other hand, Banach-Mazur determinacy principles are relatively weak on a set-theoretic level: "Every Banach-Mazur game is determined" adds no consistency strength to $\mathsf{ZF+DC}$. So the hierarchy of Banach-Mazur determinacy principles seems like a natural candidate for a set of principles "shooting off to the side" in terms of the well-understood systems of reverse mathematics (namely the big 5 + the higher $\Pi^1_n$-$\mathsf{CA}_0$s).
Question: Over $\mathsf{RCA_0}$, is there any $n$ such that $\Sigma^1_n$-$\mathsf{BM}$ implies $\Pi^1_2$-$\mathsf{CA_0}$?
I strongly suspect that the answer is negative. In fact, I conjecture that no $\Sigma^1_n$-$\mathsf{BM}$ implies $\Pi^1_1$-$\mathsf{CA_0}$. However, I do know that $\mathsf{ATR_0}$ is implied by (hence equivalent to by an old result of Steel) $\Sigma^1_1$-$\mathsf{BM}$, so I suspect that proving this stronger result might be quite hard. By contrast, a non-implication of $\Pi^1_2$-$\mathsf{CA}_0$ seems like it might just take a trick I haven't thought of.
Embarrassing admission: this question comes from a project I started way back in grad school, which despite some results languished largely due to me failing to solve this problem. I've finally given up on solving it myself, ... which on the plus side means that I hope to finish writing up what I do have fairly soon [knocks on wood].
