Note that $|F| = 2^{2019n}.$ We claim that the answer is $2019(2^{2019n} - 2^{2018n}).$ Notice that we can rewrite the summation as: $$\sum_{e \in [2019]} \sum_{(A_1, A_2, \cdots, A_n) \in F | e \in A_1 \cup A_2 \cup \cdots \cup A_n} 1.$$ Indeed, notice that for a given $e \in [2019]$, there are $2^{2018n}$ $n-$tuples in $F$ s.t. $e \notin A_1 \cup A_2 \cup \cdots \cup A_n$ (simply $2^{2018}$ ways to pick each $A_i$ to not contain $e$). This means that the inner sum is always $2^{2019n} - 2^{2018n}$, and so our answer is: $$\boxed{2019(2^{2019n} - 2^{2018n})}.$$ $\square$

Upsie this is just APMO 1998 , and my solution was same as #3