What is the topology in $\Bbb R^3$ generated by planes in $\Bbb R^3$?

In the topology generated by planes, every plane is an open set.

Once you have a topology in which every plane is an open set, you can conclude that every 1-point subset $\{x\} \subset \mathbb R^3$ is an open set, because there exist three planes whose intersection is $\{x\}$.

And once you know that every 1-point subset is open, then you can conclude that every subset is open. The topology is therefore discrete.

Regarding the basis $\mathcal B$ that is derived from the sub-basis $\mathcal A$, your formula is incorrect. The correct formula is $$\mathcal B = \{A_1 \cap \cdots \cap A_k \mid A_1,\ldots,A_k \in \mathcal A\} $$ In other words, you choose any finite indexed sequence of elements of the set $\mathcal A$, you intersect the elements of that sequence, and the result is an element of $\mathcal B$.

In the situation where $\mathcal A$ is the set of planes in $\mathbb R^3$, it follows from basic geometry that for any finite set of planes their intersection is either a plane, a line, a point, or empty. But all of that is not particularly relevant to the question of simply determining the topology generated by the set of planes; that is the discrete topology, as I showed.