Let $M$ be an $\aleph_0$-saturated structure (in a possibly uncountable language). I have a set $A \subseteq M$. Let $F$ be the family of $M$-definable subsets of $M$ containing $A$. I have the following question:
Is there a partial type $\Sigma(x)$ over a finite subset of $M$ such that every $M$-definable subset $X$ of $M$ is a superset of $\Sigma(M)$ iff $X \in F$?
That is, I want to find a set like $\bigcap F$ that is type-definable by fewer (finite) number of parameters.
When is can I find such a set, and when can I not? (For instance, it seems that I can find such a partial type for any subset $A$ in $(\Bbb Q, <)$, but that structure has very tame definable sets.)
For any structure $M$ and any $A\subseteq M$, your set $F$ includes the set $M\setminus \{b\}$ (defined by the formula $x\neq b$) for all $b\notin A$, so $\bigcap F = A$.
Now suppose we had a set $\Sigma$ satisfying the property in your question. Clearly $\Sigma(M)\subseteq \bigcap F = A$. Conversely, suppose $a\in A$. Then $M\setminus \{a\}$ is not in $F$, so $\Sigma(M)\not\subseteq (M\setminus \{a\})$, so $a\in \Sigma(M)$. So also $\Sigma(M) = A$.
So for these rather trivial reasons (we only had to look at the formula $x = y$!), you can find such a set $\Sigma$ if and only if your original set $A$ is itself type-definable over a finite set of parameters.
But it seems like you were hoping for a different kind of answer - maybe if you provided some more information about your motivation, the question could be tweaked to be less trivial.