Can there be no complexity bound on the definable elementary $V\rightarrow M$? This starts with a vaguely-recalled result (which may be false!): that if $\mathcal{U}$ is a measure on the least measurable cardinal $\kappa$, then every elementary $j: L[\mathcal{U}]\rightarrow M$ such that $(j,M)$ is first-order definable over $L[\mathcal{U}]$ with parameters from $L[\mathcal{U}]$ "comes from" iterating the elementary embedding associated to $\mathcal{U}$ itself. That's a bit vague of course, but it just occurred to me that I don't know how to whip up a situation where the opposite extreme holds:

Is it consistent with $\mathsf{MK}$ (relative to large cardinals) that for each $n$ there is a definable pair $(j,M)$ such that $j:V\rightarrow M$ is elementary but $(j,M)$ is not $\Sigma_n$-definable?

Here "definable" means "first-order definable over $V$ with set parameters." Both $j$ and $M$ are proper-class-sized objects ($M$ literally is a proper class, $j$ is a class function) so this is nontrivial; correspondingly, I'm using $\mathsf{MK}$ as my base theory here so that I can directly phrase the principle in question.
I strongly suspect that the answer to the question is yes but I don't know how to produce a model in which this holds.
 A: Yes, this is possible. If there is a proper class of measurable cardinals and $V = \text{HOD}$, then any class of ordinals $A$ is definably encoded by the iterated ultrapower $j_A : V\to M$ that hits the $\text{HOD}$-least measure on the $\alpha$-th measurable iff $\alpha\in A$. You can recover $A$ by looking at the generators of $j_A$, the ordinals $\alpha$ such that $\alpha\notin H^M(j[V]\cup \alpha)$. So if $A$ is definable but not $\Sigma_{n}$-definable, roughly the same will hold for $j_A$.
Conversely, if there is an embedding $j :V \to M$ that is not $\Delta_2$-definable from a parameter, then there is an inner model with a proper class of measurable cardinals. In fact, assuming there is no inner model with a proper class of measurable cardinals, then every elementary embedding $j:V\to M$ is the ultrapower of $V$ by a set-sized extender. (Such an ultrapower embedding is always $\Delta_2$-definable.) To see this, build the core model $K$, and note that $j$ restricts to an iterated ultrapower $i :K\to N$. If $\lambda$ is the supremum of the measurable cardinals of $K$, then all generators of $i$ must be below $i(\lambda)$, or in other words, $i$ is equal the ultrapower of $K$ by the extender of length $i(\lambda)$ derived from $i$. It follows that every ordinal is in $H^N(i[K]\cup i(\lambda)) \subseteq H^M(j[V]\cup j(\lambda))$, which means that $j$ is equal to the ultrapower of the extender of length $j(\lambda)$ derived from $j$.
