To complement Joel's answer that a positive outcome is consistently possible, let me show that the negative outcome is also possible, at least in the case of boldface definability.

For our background universe $V$, suppose that we have some second-order resources available to us, namely a global well-order and truth predicates for all class-sized structures. (I'll comment below on what this assumption entails.) This allows us to then carry out the standard argument for downward Löwenheim–Skolem. If $\mathcal N$ is a class-sized structure then we have Skolem functions for it and so can get a set-sized elementary submodel $\mathcal M$ of $\mathcal N$. In particular, $\mathcal M$ will be set-like and definable (with parameters), entailing a negative answer to your question.

Note, however, that this approach doesn't give you a negative answer for lightface definability, i.e. without parameters. Consider the case of $\mathcal N = (V,\in)$ and suppose we could find $\mathcal M$ a set-sized elementary submodel of $\mathcal N$ which is definable without parameters. But then we could define without parameters the set of true sentences in $(V,\in)$, contradicting Tarski's theorem on the undefinability of truth.

On the other hand, we can get lightface definability if we allow class quantifiers. If your global well-order is definable then, because the truth predicate is definable with class quantifiers, you get definable Skolem functions and so $\mathcal M$ is definable with class quantifiers. It's $\Delta^1_1$-definable, to be precise. Also, $\mathcal M$ is implicitly definable (without parameters), in the sense of Hamkins and Leahy - Algebraicity and implicit definability in set theory.

Let me now address what this assumption on $V$ entails, and why it doesn't apply to Joel's positive case. Having a global well-order is cheap—you can always add a generic one by class forcing, without adding any new sets. The easiest way to do this: just add a Cohen-generic subclass to $\mathrm{Ord}$. But having truth predicates comes with a cost. There's the consistency strength cost, of course, since having a truth predicate for $(V,\in)$ let's you see that ZFC is consistent. But we can say more.

**Proposition** (Essentially Krajewski): If the structure $(V,\in,\mathrm{Tr})$ satisfies ZF in the expanded language plus the assertion that $\mathrm{Tr}$ satisfies the Tarskian recursion for truth of $(V,\in)$, then $V$ contains a club of ordinals so that $V_\alpha$ is elementary in $V$.

*Proof sketch*: The point is, you can do the usual reflection argument but using $\mathrm{Tr}$ as a parameter to get the desired $V_\alpha$s, since you can express "$\mathrm{Tr}$ satisfies Tarski" as a first-order formula in the expanded language. There's a small bit more to be said about the $\omega$-nonstandard case, but it's not hard.

In particular, having truth predicates implies you have lots and lots of undefinable ordinals. So any Paris model—one whose ordinals are all definable without parameters—cannot admit a truth predicate. (Or rather—since of course every structure externally admits a truth predicate—if you add a truth predicate you destroy Replacement in the expanded language.) This includes Joel's pointwise definable $L_\lambda$.

Finally, let me note that since the $\mathcal M$ produced is a set, all we need for the counterexample is that $V$ can be extended to have the necessary second-order resources. So, for example, if your $V$ is $V_\kappa$ for some inaccessible $\kappa$, where you only look at definable classes, then you still get the counterexample. For you could expand your classes to the full powerset of $V$, apply the argument there to your definable class-sized structure $\mathcal N$, and thereby get $\mathcal M$ in $V$. But if it's in $V$, then we didn't need the extra classes to define it.