Combining Goldberg's comment and Hamkins' answer seems to work. Especially, for any inner model $M$ of ZF, we have an abstract logic $\mathcal{L}$ whose corresponding inner model $L^\mathcal{L}$ is $M$.

Consider the sublogic of $\mathcal{L}_{\infty,\omega}$ such that infinite conjunction and disjunctions are only allowed to set of formulas in $M$. In fact, $\mathcal{L}=\mathcal{L}_{\infty,\omega}^M$.

 Define $\psi_A$ for $A\in M$ as Hamkins defined: to repeat the definition, 
$$\psi_A(x):= \bigvee_{a\in A} (\forall v : v\in u\leftrightarrow \psi_a(u)).$$
Then $\psi_A(x)$ is a member of $M$ by induction on $A\in M$.

We can see that if $A\in M$, $A\subseteq V_\alpha^M$ then $$A=\{u\in V^M_\alpha \mid V^M_\alpha\models \psi_A(u)\}.$$

Hence the $\alpha$th hierarchy $L_\alpha^\mathcal{L}$ contains $V^M_\alpha$ (It can be shown by induction on $\alpha$.) Therefore $M\subseteq L^\mathcal{L}$. 
On the other hand, an inductive argument shows that the $\alpha$th hierarchy $L^\mathcal{L}_\alpha$ is a member of $M$ (we need the absoluteness of the satisfaction relation for $\mathcal{L}$ between $M$ and $V$), so $L^\mathcal{L}\subseteq M$.