In the first-order context, "reflection" of a formula $\varphi(x)$ below $\kappa$ refers to the the following situation:

There are many ordinals $\alpha<\kappa$ such that for all $a \in V_\alpha$, $V_\alpha \models \varphi(a)$ iff $V_\kappa \models \varphi(a)$.

If $\kappa$ is inaccessible, then this holds for any expansion of $V_\kappa$ in a countable language. "Many" can be taken to mean on a club set.

When we move to higher-order logic, talk of reflection usually shifts to talk of indescribability. A cardinal is $\Pi^m_n$ indescribable if for any $A \subseteq V_\kappa$ and any $\Pi_n$ sentence $\sigma$ in $(m+1)$-order logic with a predicate for $A$, if $(V_\kappa, \in, A) \models \sigma$, then there is $\alpha<\kappa$ such that $(V_\alpha,\in,A\cap V_\alpha) \models \sigma$. It is a standard fact that if $\kappa$ is measurable, then there is a measure-one set of $\alpha< \kappa$ that are $\Pi^m_n$-indescribable for every $m,n$. One can also show something stronger: If $\kappa$ is measurable, there is a measure-one set of $\alpha < \kappa$ such that if $A \subseteq V_\alpha$, then there is $\beta < \alpha$ such that $(V_\alpha,\in,A)$ and $(V_\beta,\in,A\cap V_\beta)$ have the same $\omega$-order theory.

Now this is not completely analogous to reflection because we're no longer talking about elementary substructures, but just elementarily equivalent structures, albeit with a common interpretation of a particular predicate. So my question is, what kind of large cardinal $\kappa$ is needed to get the following statement?

For any $A \subseteq V_\kappa$ and any $n \in \omega$, there are many ordinals $\alpha < \kappa$ such that $(V_\alpha,\in,A \cap V_\alpha) \prec^n (V_\kappa,\in,A)$, where $\prec^n$ means elementary in $(n+1)$-order logic.

It happens at an $\omega$-strong cardinal, but this is clearly not optimal.