On the probability of the truth of the continuum hypothesis

For any $E$ modelling ZFC and for each $n\in\omega$, we must have $(n,n)\not\in E$. Therefore $M_{ZFC}$ is contained in the cylinder set defined by $(0,0)\not\in E,\dots,(n,n)\not\in E$ which has measure $2^{-n-1}$. Therefore $M_{ZFC}$ has outer measure zero, thus is measurable and has measure zero.

Had we excluded the diagonal from $\omega\times\omega$, we can't have both $(n,m)\in E,(m,n)\in E$ for $n\neq m$, so we get infinitely many independent conditions each of measure $3/4$, so again we get outer measure zero.

Excluding foundation from ZFC makes this method not work, so it can make for a more interesting question, but perhaps we can use a similar trick.

Edit: Without foundation, let $M_n$ be the set of those models of ZFC-Foundation for which $n$ represents the empty set. We get infinitely many conditions $(m,n)\not\in E$, so $M_n$ has outer measure zero. Since $M_{ZFC-Foundation}$ is a countable union of $M_n$, it has measure zero too. Since I don't know of a reasonable variant of ZFC which doesn't prove existence of the empty set, I think that closes the case.

This answer is really just an extensive elaboration of the comments of Andreas Blass and fedja.

For any relational language $L$, let $X_L$ be the space of all $L$-structures with domain $\omega$. To determine an $L$-structure with domain $\omega$, we just need to decide, for each relation symbol $R\in L$ of arity $\text{ar}(R)$ and each tuple $a_1,\dots,a_{\text{ar}(R)}$ from $\omega$, whether or not $R(a_1,\dots,a_{\text{ar}(R)})$ holds. So $X_L$ is in bijection with the Cantor space $\prod_{R\in L}2^{\omega^{\text{ar}(R)}}$ (and the topology on $X_L$ is defined so that this bijection is a homeomorphism). Let's denote by $\mu$ the measure on $X_L$ inherited from the natural measure on the Cantor space.

The space $X_L$ also comes with an action (called the logic action) of $S_\infty$, the permutation group of $\omega$. A permutation $\sigma\in S_\infty$ acts on $X_L$ by permuating the domains of structures. That is, $$\sigma(M)\models R(a_1,\dots,a_{\text{ar}(R)})\text{ iff }M\models R(\sigma^{-1}(a_1),\dots,\sigma^{-1}(a_{\text{ar}(R)})).$$ Then $\sigma$ is an isomorphism $M\cong \sigma(M)$, and the orbit of a point $M\in X_L$ under the action of $S_\infty$ is $$\text{Iso}(M) = \{N\in X_L\mid N\cong M\}.$$ Note that the natural measure on $X_L$ is invariant for the logic action. For more on this setting, see Section 16.C of Kechris's book Classical Descriptive Set Theory.

Now the class of all finite $L$-structures is a Fraïssé class with Fraïssé limit $M_L$. It's a fact that $\mu(\text{Iso}(M_L)) = 1$ and $\text{Iso}(M_L)$ is comeager in $X_L$. So the structure $M_L$ is "generic up to isomorphism" from the point of view of both measure and Baire category.

In your question, the space on which your measure lives is $X_L$ with $L = \{E\}$, and your measure is the natural measure $\mu$. By the fact above, for any $L$-sentence $\varphi$, writing $[\varphi] = \{N\in X_L\mid N\models \varphi\}$, we have $$\mu([\varphi]) = \begin{cases} 1&\text{if } M_L\models \varphi \\ 0&\text{otherwise}\end{cases}$$ Of course, $M_L$ looks nothing like a model of set theory - it should fail to satisfy almost all the axioms, though it does satisfy extensionality.

Ok, so the natural measure $\mu$ gives measure $0$ to the models of ZFC (and the same goes for any first-order theory which is not contained in $\text{Th}(M_L)$. You might ask if there are other measures on $X_L$ which give measure $1$ to the models of ZFC. Of course, you need to impose some restrictions, since you could always pick the Dirac measure concentrating on a single point (well, assuming ZFC is consistent!).

A natural idea is to find such a measure which is still invariant for the logic action. But this is impossible, for the reason pointed out by fedja. Explicitly, suppose $\mu$ is an invariant measure on $X_L$ which gives measure $1$ to models of ZFC. For $n\in \omega$, let $Y_n$ be the set of all structures in $X_L$ such that the "set" $n$ has no elements. By invariance, $\mu(Y_n) = \mu(Y_m)$ for all $m$ and $n$, but $\mu(Y_n\cap Y_m) = 0$ for all $m\neq n$. But by countable additivity, we can't have an infinite family of almost-surely disjoint sets a probability space, all with the same positive measure.

More generally, if $\mu$ is an invariant measure on $X_L$ which gives positive measure to the set of models of a first-order theory $T$, then $T$ must have a completion with trivial definable closure: That is, there should be a model $M\models T$ such that if $\varphi(x,y)$ is a formula, where $x$ is a tuple of variables and $y$ is a singleton, and $a$ is a tuple from $M$ such that there is a unique $b\in M$ such that $M\models \varphi(a,b)$, then already $b$ is an element of the tuple $a$.

Ackerman, Freer, and Patel proved that nontrivial definable closure is actually the only obstruction to the existence of invariant measures on $X_L$ giving positive measure to a given theory. See Theorem 1.2 in this paper.

So one thing you can do is take ZFC and replace equality with an equivalence relation with infinitely many infinite classes (i.e. "blow up" each element of a model to an infinite equivalence class). This is convenient to do in the language of sets, by weakening extensionality to the axiom that if $X$ and $Y$ have exactly the same elements, then they are elements of exactly the same sets. The resulting theory ZFC' does admit invariant measures which give its set of models measure $1$ (again, assuming that ZFC is consistent). But so does (ZFC + CH)' and (ZFC + $\lnot$CH)', and I think it would be hard to make an argument to prefer the measures concentrating on (ZFC + CH)' over those concentrating on (ZFC + $\lnot$CH)' or vice versa. So unfortunately I don't think this path leads to a meaningful way to assign a "probability" to CH.