Since Turing machines have three options (accept, reject, not halt), you have a few options as to what class of languages correspond to Turing machines. The recursively enumerable sets aren't closed under complement, so your best bet is to look at recursive sets.
The recursive sets, however, aren't closed under projection. Suppose you had a correspondence between logical formulas in some language with $n$ free variables and Turing machines that always halted that take in $n$ inputs.
Consider the always-halting Turing machine $T$ that takes in two inputs $m,n$ and determines if the $n$th Turing machine halts in $m$ steps. By our supposed correspondence, this corresponds to a formula $\phi(m,n)$ in our logic. Then there also must be an always-halting Turing machine corresponding to $\exists m: \phi(m,n)$. That is, an always-halting Turing machine that takes in an $n$ and determines if the $n$th Turing machine ever halts. This contradicts the halting problem.
I consider this to rule out any proper logic equivalent to the Turing Machines. On the other hand, there are nice subsets of logics that are equivalent to the Turing Machines. For instance, by the negative result to Hilbert's Tenth Problem, the recursively enumerable sets are expressively equivalent to the existence of solutions to Diophantine equations. That is, the recursively enumerable subsets of $\mathbb{N}^m$ are exactly those that satisfy formulas of the form $\phi(y_1,\ldots,y_m) \equiv \exists x_1: \cdots \exists x_n: \phi(x_1,\ldots,x_n, y_1, \ldots, y_m)$ where $\phi$ is a quantifer-free formula in the first-order logic of $(\mathbb{N},+,\cdot,0,1)$.