We are in first order logic world.


Let $\sigma$ be a finite signature and $T$ a consistent theory of $\sigma$.


Due to Löwenheim–Skolem theorem, we can consider the $\underline{set}$ of all at most countable models of $T$ up to elementary equivalance(does not change what formulas hold).
Let $\mu$ be a nonprincipial ultrafilter of it, denote by $\Omega^T\_\mu$ the ultraproduct of
all those models with respect to $\mu$.


Clearly, the theory of $\Omega^T\_\mu$ are formulas that are true for almost all with respect to $\mu$ models of $T$.
In particular, this theory, call it $T(\Omega^T\_\mu)$, has $T$ as a subset.


My questions are:


* Does the equality $T(\Omega^T\_\mu)=T$ hold?
* Does it depend on $\mu$? Is the intersection over all $\mu$ equal to $T$?
* If they are not equal, what the difference means? Furthemore, we could consider a sequance of theories $T\subset T(\Omega^T\_\mu) \subset T(\Omega^{T(\Omega^T\_\mu)}\_{\mu'})\subset\ldots$. Does it stabilize?
* Does $T(\Omega^T\_\mu)$ complete $T$ in any nice sense?


I suspect that all those questions are classical and already answered, but I do not know where
to look for them. Just before pandemia I have finished my first course on mathematical logic and I have been wondering about this object since then.



The theory of a model of a theory that knows all formulas true in almost all its models
In this answer, I will basically repeat things that zeb said in their nice answer, but arranged differently (in a way that might or might not be more clear). 


Let $\text{Mod}\_{\leq \aleph\_0}(T)$ be the "set" of all at most countable models of $T$. Your question is a bit ambiguous about what this means - I'll come back to that later. Morally, picking an ultrafilter on $\text{Mod}\_{\leq \aleph\_0}(T)$ is the same as picking a completion of $T$. 


In one direction, if you pick an ultrafilter $\mu$ on $\text{Mod}\_{\leq \aleph\_0}(T)$, you can take the complete theory of the ultraproduct, $T(\Omega^T\_\mu)$ in your notation. This is a complete theory, and it contains $T$, so it's a completion of $T$. 


In the other direction, given a completion $T\subseteq T'$, for each sentence $\varphi\in T'$, consider the set of models $M\_\varphi = \{M\in \text{Mod}\_{\leq \aleph\_0}(T)\mid M\models \varphi\}$. Then $F\_{T'} = \{M\_\varphi\mid \varphi\in T'\}$ has the finite intersection property, so we can extend it to an ultrafilter. And moreover for any ultrafilter $\mu$ extending $F\_{T'}$, we have $T(\Omega^T\_\mu) = T'$, by Łoś's Theorem. 


But now there's a subtlety here: You want $\mu$ to be a non-principal ultrafilter.


If by $\text{Mod}\_{\leq \aleph\_0}(T)$, you mean a set of isomorphism representatives for the models of $T$, so this set contains just one copy of each model of $T$ up to isomorphism, then if $T'$ has only one (at most countable) model (i.e. if it's the theory of a finite structure, or if it's $\aleph\_0$-categorical), then the ultrafilter $\mu$ you get by the above construction will be principal. That is, picking a non-principal ultrafilter on $\text{Mod}\_{\leq \aleph\_0}(T)$ picks a completion of $T$, but we never get a completion that is $\aleph\_0$-categorical or the theory of a finite structure. Worse, if $T$ already has only finitely many (at most countable) models up to isomorphism, then there is *no* non-principal ultrafilter on $\text{Mod}\_{\leq \aleph\_0}(T)$ - this corresponds to the fact that $T$ has no completion with more than one (at most countable) model. 


These issues go away if $\text{Mod}\_{\leq \aleph\_0}(T)$ contains infinitely many copies of each model up to isomorphism (e.g. if it's more like the proper class of *all* at most countable models of $T$). Then when we extend $F\_{T'}$ to an ultrafilter $\mu$, we can always make sure $\mu$ is non-principal. 


So the answer to your questions are: 


(1) $T(\Omega\_\mu^T) = T$ if and only if $T$ is already complete.


(2) Yes, it definitely depends on $\mu$ (unless $T$ is complete). If you remove the restriction that $\mu$ be non-principal, the intersection of all theories of the form $T(\Omega^T\_\mu)$ is $T$. If you require $\mu$ to be non-principal, this intersection is the set $T\_{\text{cof}}$ of all sentences which are in all but finitely many models in $\text{Mod}\_{\leq \aleph\_0}(T)$. If $\text{Mod}\_{\leq \aleph\_0}(T)$ contains infinitely many copies of each model up to isomorphism (e.g. if this "set" is actually the proper class of all at most countable models of $T$), then again $T\_{\text{cof}} = T$. 


(3) The "difference" consists of moving from $T$ to a completion. The chain stabilizes after the first step, i.e. $T(\Omega\_\mu^T) = T(\Omega\_{\mu'}^{T(\Omega\_\mu^T)})$, since $T(\Omega\_\mu^T)$ is already complete.


(4) Yes, $T(\Omega\_\mu^T)$ is a completion of $T$. 


