I think that for any given theory $T$, the statement in the question should be true.
My attempt of proof:
Let us fix a theory $T$ and a cardinal $\kappa$, we introduce variables $\{ z_\alpha : \alpha<\kappa \}$. We would like to construct a type $p$ in this variables such that $(a_\alpha)_{\alpha<\kappa}$ satisfy $p$ if and only if $\{ a_\alpha : \alpha<\kappa \}$ is the universe of a model of $T$.
I first looked at sentences with only one quantifier. Let $\varphi(x)$ be q-f. If $T$ implies $\forall x \varphi(x)$, then we add to $p$ the formula $\varphi(z_\alpha)$ for every $\alpha$. But if we are in the case $\exists x \varphi(x)$ I don't really know how to proceed. I would like to add something equivalent to $\bigvee_{\alpha<\kappa} \varphi(z_\alpha)$.
What you ask for, literally, is impossible, as in the example given by Noah in the comments --- in a type $p(\bar x)$ in infinitely many variables, you cannot assert in a first order way that any one of the variables in $\bar x$ represents the constant $c$.
What is possible is finding a type $p((x_\alpha)_{\alpha<\lambda})$ over $\emptyset$ such that the elementary submodels (of cardinality $\leq \lambda$ or $=\lambda$) are exactly the sets enumerated by realisations of $p$ (usually non-uniquely and non-injectively!).
Note that I will assume $\lambda\geq \lvert T\vert$. I suspect this should not be hard to extend to small models of large theories, but there will likely be some technicalities involved that I would rather avoid.
For a given set $M$ to be an elementary submodel, it is sufficient (and necessary) to know that it satisfies the Tarski-Vaught test, i.e. that for any formula $\varphi(x,\bar y)$ and any tuple $\bar b\in M$ such that $\models\exists x\varphi(x,\bar b)$, there is an $a\in M$ such that $\models \varphi(a,\bar b)$.
You are given a tuple $I = (x_\alpha)_{\alpha<\lambda}$ of variables, where $\lambda$ is an infinite cardinal $\geq \lvert T\rvert$.
To ensure that the Tarski-Vaught test is satisfied, you want to have, for each formula $\varphi(x,\bar y)$ and each tuple $\bar y\subseteq I$, there is some variable $z=z_{\bar y, \varphi}\in X$, such that your type has the formula $\tilde\varphi(\bar y, z_{\bar y,\varphi})=(\exists x \varphi(x,\bar y))\rightarrow \varphi(z_{\bar y,\varphi}, \bar y)$.
Now, in order to ensure that this is satisfiable, you need to be careful with your choice of $z_{\bar y,\varphi}$. The easiest way to do this is to first enumerate all pairs $(\bar y, \varphi)$ as a $\lambda$-sequence $(\bar y_\alpha,\varphi_\alpha)_{\alpha<\lambda}$ (this is why I wanted to assume that $\lambda\geq \lvert T\rvert$ --- otherwise, there would be too many formulas). To ensure that there is never any clash, you should choose $z_\alpha= z_{\bar y_\alpha,\varphi_\alpha}$ in such a way that $z_\alpha\neq z_\beta$ for $\beta<\alpha$ and $z_\alpha\notin \bar y_\beta$ for any $\beta\leq \alpha$. This is standard book-keeping. Given this, you can recursively find a realisation of this type (in any model of $T$, in fact), and it is not hard to see that it enumerates a submodel (by the Tarski-Vaught test).
Now, every realisation of this type will enumerate an elementary submodel. However, it is not obvious (to me) that every elementary submodel of cardinality $\leq \lambda$ has an enumeration satisfying this type. In order to have that, you need to ensure that there are $\lambda$-many variables which are not $z_\alpha$ for any $\alpha$. This is, again, straightforward book-keeping. Then given any elementary submodel, you can enumerate the model by the "un-assigned" variables, and then extend the enumeration to all variables in accordance with the Tarski-Vaught test.
(In order to ensure that you do not get any models of smaller cardinality than $\lambda$, you should additionally inject a requirement that $\lambda$ many variables refer to different elements, but that is also quite straightforward.)