Trakhtenbrot's theorem states that given a sentence $\phi$ in a finite vocabulary (M. Viswanathan, "Finite Model Theory", notes, 2018), the problem of whether $\phi$ is satisfied in a finite model is undecidable. Is the problem of whether a sentence $\phi$ in a given first-order logic is satisfiable in a computable model recursively enumerable? To be more specific, if $\phi$ ranges over formulas in a first-order logic with relation symbols $(R_i)_{i<k}$, is the problem of whether $\phi$ is satisfied in some model $(\mathbb N,(R_i)_{i<k})$, with $R_i$ computable for all $i<k$, a recursively enumerable decision problem?
There is an effective version of the completeness theorem stating that a decidable theory has a decidable (and therefore computable) model, on page 18 of Harizanov's Chapter 1 ("Pure Computable Model Theory") of the book Handbook of Recursive Mathematics, this presentation allows infinitely many relation symbols and requires uniform computability of their interpretations. On the other hand there should be some finitely axiomatizable theories with no computable models, for example there should be no computable relation $R$ on $\mathbb N$ such that $(\mathbb N,R)$ satisfies Kripke-Platek set theory, by effectively interpreting a nonstandard model of Peano arithmetic and applying Tennenbaum's theorem.
No, it's not. For example, working in the language of arithmetic (appropriately relationalized) look at sentences of the form $\alpha\wedge\varphi$ where $\alpha$ is a fixed sentence axiomatizing a large enough fragment of arithmetic for Tennenbaum's theorem to apply. Then $\alpha\wedge\varphi$ has a computable model iff $\varphi\in\mathsf{TA}$. This is not even arithmetical, let alone c.e.
On the other hand, the above problem is "locally arithmetical" in the sense that for each fixed $n$ the set of (Godel numbers of) $n$-quantifier sentences which are in $\mathsf{TA}$ is arithmetical. We can't do significantly better (worse?) than this, since the set of $n$-quantifier sentences which have computable models is $\Sigma_{\max\{n,3\}+1}$.