For some non-standard model $M \models PA$, I recently proved that there is some $a \in M$ such that for all quantifier-free formulas $\varphi(x)$, if $M \models (\exists \, x) \varphi(x)$, then $M \models (\exists \, x<a) \varphi(x)$.
This can be done in the following way: for $b \in M$ non-standard and $\left\{ \varphi_{n}(x) \right\}_{n < b}$ which are quantifier-free formulas in $M$ with indices less than $b$ define a function $f(n)$ which returns $x$ for $n$ the least solution to $\varphi_{n}(x)$ and $0$ else. Then $f$ is definable and moreover $\Delta_{2}$. Now consider a bound $a$ on the range of $f$ on $\mathbb{N} \subseteq [0,b]$ such that $PA \vdash B \Sigma_{n}$. This is: $PA \vdash \forall \,a [(\forall \,x \leq a)(\exists \, y) \varphi(x,y) \rightarrow \exists \, b(\forall \, x \leq a)(\exists \, y \leq b)\varphi(x,y)]$. Thus, a non-standard model has a non-standard bound on the range of $f$. Thus, for every quantifier-free $\varphi(x)$, these appear as $\varphi_{n}(x)$ for standard $n$ such that $M \models (\exists \, x<a) \varphi(x)$.
I'm pretty sure this works, but if not any correction would be appreciated. My question is: would this proof change if we consider no quantifier-free formulas, but rather, say, $\Sigma_{n}$ formulas? Are there any additional obstructions?
Nope, this proof works fine for any class of formulas with an outermost existential quantifier for which a truth predicate is defined - e.g. any of the classes $\Sigma_n$. We get a function $f$ sending the $n$th such formula to its witness if there is one, and to $0$ otherwise; this $f$ is definable at the $\Sigma_{n+17}$-level (more optimal bounds can be found :P), and so PA proves that the $f$-image of the set $[0, b]$ is bounded above for any nonstandard $b$, at which point we're done.
That said, the sentence "Now consider a bound $a$ on the range of $f$ on $\mathbb{N}\subseteq [0, b]$ such that $PA\vdash B\Sigma_n$" is garbled: I think you mean "$B\Sigma_2$" rather than "$B\Sigma_n$", and you should say "and recall that" instead of "such that" (since $PA$ just outright proves $B\Sigma_2$, it doesn't depend on the choice of $a$).