You say that "To find $\dim (A_1+A_2+⋯+A_n)$ it seems we can use inclusion exclusion", so I want to point out that while for $n=2$ the inclusion-exclusion formula $ \dim (A+B) = \dim A +\dim B - \dim A\cap B$ is true, it fails for $n=3$: in general $\dim (A+B+C) \neq \dim A+\dim B + \dim C - \dim(A\cap B) - \dim(B\cap C) - \dim (A \cap C) + \dim(A\cap B \cap C).$ Look at the example of three distinct lines in $\mathbb{R}^2$.
The problem is that subspaces of a vector space don't form a distributive lattice, in other words $A \cap (B + C) \neq A \cap B + A \cap C$. Repeatedly using the $n=2$ formula will give you an expresion for $\dim (A+B+C)$, but not just in terms of the terms appearing on the right of the displayed equation above.