You get $2^\kappa$ many models for any uncountable $\kappa$. This follows immediately from the fact that $\mathsf{ZFC}$ (trivially) defines an infinite linear order, and this is all that's needed to ensure unstability.
(The same argument shows that already much weaker theories, such as $\mathsf{PA}$, are unstable.)
This gives the result for $\kappa$ uncountable. For $\kappa$ countable you also have the maximum number of nonisomorphic models of $\mathsf{ZFC}$. To see this, use the incompleteness theorem to show that you can recursively label the nodes of the complete binary tree as $T_s$, $s\in 2^{<\omega}$, so that $T_\emptyset=\mathsf{ZFC}$ (or $\mathsf{PA}$, if you prefer), $T_s$ is consistent for each $s$, $T_s\subsetneq T_t$ for $s\subsetneq t$, and each $T_{s{}^\frown\langle i\rangle}$, for $i=0,1$, is obtained from $T_s$ by adding a single new axiom (that depends on $s$, of course). Now, for each $x\in 2^\omega$, let $T_x$ be any consistent complete theory extending $\bigcup_n T_{x\upharpoonright n}$. These are $2^{\aleph_0}$ pairwise incompatible theories, all extending $\mathsf{ZFC}$ (or $\mathsf{PA}$), and all having countable models. (Examples such as the theory of $(\mathbb Q,<)$ show that the argument must be different for countable models.)
The paragraph above shows that there are $2^{\aleph_0}$ incompatible extensions of $\mathsf{ZFC}$, and therefore $2^{\aleph_0}$ non-isomorphic countable models of $\mathsf{ZFC}$. The result is also true for a fixed complete extension $T$, but the argument seems harder. A proof follows from the following, but most likely there are easier approaches:
Consider first the paper
Ali Enayat. Leibnizian models of set theory, J. Symbolic Logic, 69 (3), (2004), 775–789. MR2078921 (2005e:03076).
In it, Enayat defines a model to be Leibnizian iff it has no pair of indiscernibles. He shows that there is a first order statement $\mathsf{LM}$ such that any (consistent) complete extension $T$ of $\mathsf{ZF}$ admits a Leibnizian model iff $T\vdash\mathsf{LM}$. He also shows that $\mathsf{LM}$ follows from $\mathrm{V}=\mathsf{OD}$, and that any (consistent) complete extension of $\mathsf{ZF}+\mathsf{LM}$ admits continuum many countable nonisomorphic Leibnizian models.
Now consider the paper
Ali Enayat. Models of set theory with definable ordinals, Arch. Math. Logic, 44 (3), (2005), 363–385. MR2140616 (2005m:03098).
In it, Enayat defines a model to be Paris iff all its ordinals are first order definable within the model. He shows that any (consistent) complete extension of $\mathsf{ZF}+\mathrm{V}\ne\mathsf{OD}$ admits continuum many countable nonisomorphic Paris models.
These two facts together imply the result you are after (and more, of course). An earlier result of Keisler and Morley, that started the whole area of model theory of set theory, shows that any countable model of $\mathsf{ZFC}$ admits an elementary end-extension. (This fails for uncountable models.) It may well be that an easy extension of this is all that is needed to prove the existence of continuum many non-isomorphic countable models of any fixed (consistent) $T\supset\mathsf{ZFC}$, but I do not see right now how to get there. The Keisler-Morley theorem alone does not seem to suffice, in view of Joel Hamkins's beautiful answer to this question.
So the question is: what am I missing in the assertion that ZFC is relatively consistent with the absence of set models? Does this just mean that some models of ZFC have no elements which are inner models?
Yes (though "inner model" has a technical meaning which I'm pretty sure is not what you intend here). To be precise, it means that ZFC (assuming it is consistent) cannot prove there exists a model of ZFC (here by model I will always mean set model). Therefore, by the completeness theorem, assuming ZFC is consistent, there exists a model $(M,\epsilon)$ of ZFC such that there does not exist any $m\in M$ such that $(M,\epsilon)\vDash\text{"$m$ is a model of ZFC"}$, where "$m$ is a model of ZFC" is the usual statement in the language of set theory encoding the statement that $m$ is a model of ZFC.
(Note that such a model $(M,\epsilon)$ cannot think that ZFC is consistent, since if it did, then by the completeness theorem internalized to $(M,\epsilon)$ there would be some element $m\in M$ which $(M,\epsilon)$ thinks is a model of ZFC. So in this setup, ZFC really is consistent, but $(M,\epsilon)$ believes it is inconsistent, and has a nonstandard natural number that encodes a "proof" of a contradiction from ZFC.)
Best Answer
There are some subtleties around class-sized models, but any reasonable approach to handling them will give an affirmative answer to your question if you restrict attention to "locally set-like" models, i.e. models $\mathcal{M}$ for whichevery class of the form $$\{n\in\mathcal{M}: n\in^\mathcal{M}m\}$$ for $m\in\mathcal{M}$ is actually a set.
To see this, we just follow the usual "quasi-categoricity" argument for the set-sized case. By second-order Foundation, $\mathcal{M}$ must be well-founded; by second-order Powerset, after taking a Mostowski collapse (permitted by the well-foundedness we've just established) we must have $V_\alpha^\mathcal{M}=V_\alpha$ for each ordinal $\alpha$. And since $\mathcal{M}$ is class-sized, that leaves only one candidate: $\mathcal{M}=V$.
But this analysis breaks down if we allow $\mathcal{M}$ to not be locally set-like. In particular, the set-like part of $\mathcal{M}\models^{class}\mathsf{ZFC_2}$ will always be (isomorphic to) $V$ itself, but it may be a "proper top extension" of $V$. On the other hand, no such $\mathcal{M}$ will be first-order definable in $V$, so if you restrict attention to definable class models we recover categoricity.
(Note that there are $V$-definable well-founded non-locally-set-like structures; e.g. take $\mathit{Ord}$ with the order that puts $0$ above everything else and otherwise is ordered as usual. So the definability barrier here is nontrivial.)