Given the sets
$$A_{1} = \{ a_{11}, a_{12}, a_{13}, a_{14}, ..., a_{1(n-1)} \}$$
$$A_{2} = \{ a_{11}, a_{21}, a_{22}, a_{23}, ..., a_{2(n-2)} \}$$
$$A_{3} = \{ a_{12}, a_{21}, a_{31}, a_{32}, ..., a_{3(n-3)} \}$$
$$A_{4} = \{ a_{13}, a_{22}, a_{31}, a_{41}, ..., a_{4(n-4)} \}$$
$$ ... $$
$$A_{n} = \{ a_{1(n-1)}, a_{2(n-2)}, a_{3(n-3)}, a_{4(n-4)}, ..., a_{(n-1)1} \}$$
If we choose the first $k$ sets from $A_1,A_2,...,A_n$, then how many possible sets can be created by combining the elements of $A_1, A_2,..., A_k$ such that at least one element is in $A_1$, at least one element is in $A_2$, ..., and at least one element is in $A_k$?
In other words, let $B_i \subseteq A_i$ for $i=1,2,...,n$. For a given $n \in \mathbb{Z}^{\ge 2}$, I am seeking a function $f: \{ 2,3,...,n \} \to \mathbb{N}$ where $f(k)$ is the number of distinct sets given by $\cup_{j=1}^{k} B_j$ such that $\cup_{j=1}^{k} B_j$ contains at least $1$ element from $A_j$ for $j=1,2,...,k$.
As a starting point, I determined that if $A_1, A_2,..., A_k$ were disjoint, then the answer would be
$$\sum_{j_1=1}^{n-1} \sum_{j_2=1}^{n-1} ... \sum_{j_k=1}^{n-1} {n-1\choose j_1}{n-1\choose j_2}...{n-1\choose j_k}$$
However, although the elements within each set are distinct, no two sets are disjoint, and this makes the problem difficult to wrap my mind around. I have observed the following pattern for how the sets intersect, though: given the choice of $A_1, A_2,..., A_k$, the set $A_m$ contains exactly $m-1$ elements $x_1,x_2,...,x_{m-1}$ such that $x_1 \in A_1$, $x_2 \in A_2$, ..., $x_{m-1} \in A_{m-1}$, where $2 \le m \le k$. Furthermore, this is true no matter how we order the sets $A_1,A_2,...,A_n$.
Any help is appreciated. Thank you!