The Tennenbaum phenomenon is amazing, and that is totally correct, but let me give a direct proof using the idea of computable inseparability.
Theorem. There is no computable model of ZFC.
Proof: Suppose to the contrary that M is a computable model of ZFC. That is, we assume that the underlying set of M is ω and the membership relation E of M is computable.
First, we may overcome the issue you mention at the end of your question, and we can computably get access to what M thinks of as the
nth natural number, for any natural number n. To see this, observe first that there is a particular natural number z, which M believes
is the natural number 0, another natural number N, which M believes to the set of all natural numbers, and another natural number s, which M
believes is the successor function on the natural numbers. By decoding what it means to evaluate a function in set theory using ordered pairs, We
may now successively compute the function i(0)=z and i(n+1) = the unique number that M believes is the successor function s of i(n). Thus,
externally, we now have computable access to what M believes is the nth natural number. Let me denote i(n) simply by n. (We
could computably rearrange things, if desired, so that these were, say, the odd numbers).
Let A, B be any computably inseparable sets. That is, A and B are disjoint computably
enumerable sets having no computable separation. (For example, A is the set of TM programs halting with output 0 on input 0, and B is the set of
programs halting with output 1 on input 0.) Since A and B are each computably enumerable, there are programs p0 and p1 that
enumerate them (in our universe). These programs are finite, and M agrees that p0 and p1 are TM programs that
enumerate a set of what it thinks are natural numbers. There is some particular natural number c that M thinks is the set of natural numbers
enumerated by p0 before they are enumerated by p1. Let A+ = { n | n E c }, which is the set
of natural numbers n that M thinks are enumerated into M's version of A before they are enumerated into M's version of B. This is a computable
set, since E is computable. Also, every member of A is in A+, since any number actually enumerated into A will be seen by M to have
been so. Finally, for the same reason, no member of B is in A+, because M can see that they are enumerated into B by a (standard)
stage, when they have not been enumerated into A. Thus, A+ is a computable separation of A and B, a contradiction. QED
Essentially this argument also establishes the version of Tennenbaum's theorem mentioned by Anonymous, that there is no computable nonstandard
model of PA. But actually, Tennenbaum proved a stronger result, showing that neither plus nor times individually is computable in a nonstandard model of PA. And this takes a somewhat more subtle argument.
Edit (4/11/2017). The question was just bumped by another edit, and so I thought it made sense to update my answer here by mentioning a generalization of the result. Namely, in recent joint work,
we prove that not only is there no computable model of ZFC, but also there is no computable quotient presentation of a model of ZFC, and indeed there is no c.e. quotient presentation of such a model, by an equivalence relation of any complexity. That is, there is no c.e. relation $\hat\in$ and equivalence relation $E$, of any complexity, which is a congruence with respect to $\hat\in$, such that the quotient $\langle\mathbb{N},\hat\in\rangle/E$ is a model of ZFC.
The least ordinal not in any transitive model of ZFC can also be described as the supremum of the heights of transitive models of ZFC. It is natural here to consider the class S consisting of all ordinals λ for which there is a transitive model of ZFC of height λ. Thus, the ordinal of your title, when it exists, is simply the supremum of the countable members of S. There are a number of relatively easy observations:
If M is a transitive model of ZFC, then so is LM, the constructible universe as constructed inside M, and these two models have the same height. Thus, one could equivalently consider only models of ZFC + V = L.
It is relatively consistent with ZFC that S is empty, that is, that there are no transitive models of ZFC. For example, the least element of S is the least α such that Lα is a model of ZFC. This is sometimes called the minimal model of ZFC, though of course it refers to the minimal transitive model. It is contained as a subclass of all other transitive models of ZFC. The minimal model has no transitive models inside it, and so it believes S to be empty.
The least element of S (and many subsequent elements) is Δ12 definable in V. This is because one can say: a real codes that ordinal iff it codes a well-ordered relation and there is a model of that order type satisfying that no smaller ordinal is in S (a Σ12 property), also iff every well-founded model of ZFC has ordinal height at least the order type of the ordinal coded by z (a Π12 property).
If S has any uncountable elements, then it is unbounded in ω1. The reason is that if Lβ satisfies ZFC and β is uncountable, then we may form increasingly large countable elementary substructures of Lβ, whose Mostowski collapses will give rise to increasingly large countable ordinals in S.
In particular, if there are any large cardinals, such as an inaccessible cardinal, then S will have many countable members.
If 0# exists, then every cardinal is a member of S. This is because when 0# exists, then every cardinal κ is an L-indiscernible, and so Lκ is a model of ZFC. Thus, under 0#, the class S contains a proper class club, and contains a club in every cardinal.
S is not closed. For example, the supremum of the first ω many elements of S cannot be a member of S. The reason is that if αn is the nth element of S, and λ = supn αn, then there would be a definable cofinal ω sequence in Lλ, contrary to the Replacement axiom.
S contains members of every infinite cardinality less than its supremum. If β is in S, then we may form elementary substructures of Lβ of any smaller cardinality, and the Mostowski collapses of these structures will give rise to smaller ordinals in S.
If β is any particular element of S, the we may chop off the universe at β and consider the model Lβ. Below β, the model Lβ calculates S that same as we do. Thus, if β is a limit point of S, then Lβ will believe that S is a proper class. If β is a successor element of S, then Lβ will believe that S is bounded. Indeed, if β is the αth element of S, then in any case, Lβ believes that there are α many elements of S.
If S is bounded, then we may go to a forcing extension V[G] which collapses cardinals, so that the supremum of S is now a countable ordinal. The forcing does not affect whether any Lα satisfies ZFC, and thus does not affect S.
Reading your question again, I see that perhaps you meant to consider a fixed M, rather than letting M vary over all transitive models. In this case, you will want to look at fine-structural properties of this particular ordinal. Of course, it exhibits many closure properties, since any construction from below that can be carried out in ZFC can be carried out inside M, and therefore will not reach up to ht(M).
Best Answer
No. Because the construction of the inner model $L$ is absolute for transitive models of ZFC the ordinal heights of transitive models of ZFC are precisely the ordinals $\alpha$ such that $L_\alpha$ is a model of ZFC. If $\alpha_0,\alpha_1$ are the first two ordinals such that $L_{\alpha_0}$ and $L_{\alpha_1}$ are models of ZFC, then $L_{\alpha_1}$ is a model of ZFC in which there is a transitive model of ZFC, namely $L_{\alpha_0}$, and every transitive model of ZFC has rank $\alpha_0$ and is therefore bounded in size by $|V_{\alpha_0}|$.