https://mathoverflow.net/q/93242/7206 shows that assuming $\sf BPI$ we have that every two bases have the same cardinality. This means that $BS(V)$ is either empty, or a singleton.

So in a model of $\sf ZF+\lnot AC+BPI$ we have that $\frak BS$ is singletons and the empty set, which is definitely closed under finite (or otherwise) intersections.

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As for a general structure, note that there is a strong restriction on the elements of $BS(V)$, when $V$ is a vector space over $F$. Namely, whenever $B\in BS(V)$ then $$|V|=|[F\times B]^{<\omega}|.$$

This means that there are bijections between the various $[F\times B]^{<\omega}$ sets. So even if $BS(V)$ has several different elements, they are all bounded below $|V|$.

So over a fixed field, if $B$ is a basis for a vector space of one cardinality, and $B'$ is a vector space of a different cardinality, then $\{B,B'\}$ cannot possible be a member of $\frak BS$.