Here's an example showing that in general, for pregeometries arising in model theory, you can't characterize the dimension of a union of a chain of models just in terms of the dimensions of the models. In other words, it matters how the models embed into each other.
Consider the theory of a single equivalence relation $E$ with infinitely many infinite classes, and define $\dim(M) = |M/E|$.
First union: Let $M_0$ be the unique countable model of this theory up to isomorphism. For every countable ordinal $\alpha$, let $M_{\alpha+1}$ be the elementary extension of $M_\alpha$ obtained by adding a single new equivalence class with countably many elements. For a limit ordinal $\lambda$, let $M_\lambda = \bigcup_{\alpha<\lambda} M_\alpha$. Then $\dim(M_\alpha) = \aleph_0$ for all $\alpha<\omega_1$, but $\dim(M_{\omega_1}) = \aleph_1$,
Second union: Let $N_0 = M_0$, and pick an equivalence class $C$. This time, for every $\alpha$, let $N_{\alpha+1}$ be the elementary extension of $N_\alpha$ obtained by adding a single new element to $C$. As before, take unions as limit stages. Then $\dim(N_\alpha) = \aleph_0$ for all $\alpha<\omega_1$, and also $\dim(N_{\omega_1}) = \aleph_0$.
Now I need to convince you that this dimension function arises in a standard way from a pregeometry studied in model theory. In stability theory, there's the notion of a regular type: a stationary type which is orthogonal to all of its forking extensions. The key point is that if $p(x)$ is a regular type (let's say over $\emptyset$ for simplicity), then forking dependence gives rise to a pregometry on the realizations of $p$ via the closure operator $\mathrm{cl}(A) = \{b\models p(x)\mid \text{tp}(b/A) \text{ forks over }\emptyset\}$.
In my example, the theory is stable, the unique $1$-type is a regular type, and forking is governed by the equivalence relation $E$, so we get a pregometry on the whole model with closure operator $\mathrm{cl}(A) = \bigcup_{a\in A} [a]_E$, where $[a]_E$ is the $E$-class of $a$. And the induced dimension function is $\dim(M) = |M/E|$.
Well, maybe you don't like this kind of pregeometry, and you only want to consider the kind you meet more often in model theory, namely pregeometries induced by the $\text{acl}$ operator. That's fine, but then the dimensions are only interesting for models that are at most the size of the language (so only countable models if the language is countable).
Indeed, suppose $T$ is a theory such that $\text{acl}$ induces a pregometry on every model of $T$, and let $M\models T$ with $|M| > |L|$. Since $|\text{acl}(A)| = \max(|A|,|L|)$ for all $A\subseteq M$, any basis for $M$ must have cardinality $|M|$, and $\dim(M) = |M|$.
In this case, the answer to your question about unions is an easy yes.
Added in edit: You might also decide that you're only interested in closure operators with the property that when $A\subseteq M\prec N$, $\text{cl}(A)$ in $M$ equals $\text{cl}(A)$ in $N$, i.e. closures don't grow in elementary extensions. This is the case for $\text{cl} = \text{acl}$, and it would salvage the proof in your answer that $\dim$ takes unions of chains to sums, since if $N$ is a proper elementary extension of $M$, the closure of a basis for $M$ is contained in $M$, and we need at least one new element to form a basis for $N$. But we actually don't get anything beyond $\text{acl}$ under this assumption.
Indeed, suppose $\text{cl}$ satisfies the condition above, and look at $A\subseteq M$. Embed $M$ in a large monster model $\mathbb{M}$. Then $\text{cl}_M(A) = \text{cl}_{\mathbb{M}}(A)$. In fact, for any $A\subseteq N\prec \mathbb{M}$, we have $\text{cl}_N(A) = \text{cl}_{\mathbb{M}}(A)$, so $\text{cl}_{\mathbb{M}}(A)\subseteq N$. But $\bigcap\{N\mid A\subseteq N\prec \mathbb{M}\} = \text{acl}(A)$, so $\text{cl}(A)\subseteq \text{acl}(A)$.
If you're interested in axiomatizing dimension functions, you might want to look this paper, which gives a number of equivalent axiom systems for infinite matroids. In particular, look at the axioms in terms of rank functions. Their rank functions take values in $\mathbb{N}\cup \{\infty\}$, but you might as well be in this situation if you're thinking about $\text{acl}$ pregometries ($\dim(M) = \infty$ means $\dim(M) = |M|$).