Here is a question on recursively saturated model (2.4.14 in Chang and Keisler's Model Theory).
Let $\mathcal{U}$ be a recursively saturated model for recursive language, and let $$\{\phi_m(x_1,\cdots,x_n,y_1,\cdots,y_n):m<\omega\} $$ be a r.e. set of formulas. Suppose that for each $m<\omega$, $$ \mathcal{U}\models \phi_{m+1}\to \phi_m\quad\text{and}\quad \mathcal{U}\models \forall x_1\exists y_1\cdots\forall x_n\exists y_n\:\phi_{m} $$ Prove that $$ \mathcal{U}\models \forall x_1\exists y_1\cdots\forall x_n\exists y_n\bigwedge\limits_{m<\omega}\phi_{m} $$
Here's a proof of a simpler statement: we replace "recursively saturated" by "(countably) saturated," drop "r.e.," and set $n=1$.
Under the given assumptions, we want to show $\mathcal{U}\models\forall x\exists y\bigwedge_{i<\omega}\varphi_i(x,y)$. So fix $a\in\mathcal{U}$.
For each $i<\omega$ we have by our second assumption that there is some $b_i^a\in\mathcal{U}$ such that $\varphi_i(a,b_i^a)$ holds. Moreover, by our first assumption we have in fact that $\bigwedge_{j\le i}\varphi_j(a,b_i^a)$ holds. This means that the partial $\{a\}$-type $$\{\varphi_i(a,x): i<\omega\}$$ is finitely satisfiable in $\mathcal{U}$, hence by saturation is realized by some $b^a_\infty\in\mathcal{U}$.
The map $a\mapsto b^a_\infty$ witnesses the truth of $\forall x\exists y\bigwedge_{i<\omega}\varphi_i(x,y)$ in $\mathcal{U}$.
From this it's not hard to deduce the desired result; shifting to the recursion-theoretic version basically amounts to just checking the relevant complexities, and generalizing to arbitrary $n$ is an easy induction.