Tuples of sets $A_1, ... , A_k \in \mathcal{P}([n])$ with $\bigcup_{i=1}^{k}{A_i}=[n]$ and $A_i \neq A_j$, $1 \le i \lt j \le k$ To generalize this question I would like to count how many tuples of sets $A_1, ... , A_k \in \mathcal{P}([n])$ with $\bigcup_{i=1}^{k}{A_i}=[n]$ and $A_i \neq A_j$, for any $i,j$ such that $1 \le i \lt j \le k$, are there.
The case $k=2$ has been solved in the linked question and the result is $\frac{3^n-1}{2!}$.
The case $k=3$ can be stated as $\frac{7^n-1}{3!}-\frac{3^n-1}{2!}$ and a computer test confirms the result.
For case $k = 4$ I would have expected something like $\frac{15^n-1}{4!}-\frac{7^n-1}{3!}+\frac{3^n-1}{2!}$, but the first addendum is not an integer. The values computed (hopefully correctly) for $n=1,2,3,4,5,6$ are $0,1,67,1546,27550,445531$.
To complete with computing data, the case $k = 5$ gives $0,0,56,4144,180096,6480656$, for $k = 6$ we have $0,0,28,7896,866432,69656776$, for $k = 7$: $0,0,8,11408,3308736,601192496$.
Any hint? If possible, for the general case, but especially for $k = 4$. Thank you.

Another, non direct, way to answer the linked question is by using binomial inversion for the numerical part and using an inclusion-exclusion argument, so that if $$\binom{2^m}{2}=\sum_{n=0}^m {m \choose n}\frac{3^n-1}{2},$$ then $$\frac{3^m-1}{2}=(-1)^m\sum _{n=0}^m\binom{m}{n}(-1)^n\binom{2^n}{2}.$$ In this way, I think you are looking for
$$(-1)^m\sum _{n=0}^m\binom{m}{n}(-1)^n\binom{2^n}{k}.$$
which agrees with your computation. You can check this combinatorially using inclusion-exclusion by considering which element is missing in your decomposition.
Edit:
Consider then
$$(-1)^m\sum _{n=0}^m\binom{m}{n}(-1)^n\binom{2^n}{k}=\binom{2^{m}}{k}-\sum _{n=1}^{m}\binom{m}{n}(-1)^{n-1}\binom{2^{m-n}}{k},$$
this corresponds to doing $$\left |\mathcal{A}\setminus \bigcup _{\ell=1}^m\mathcal{A}_{\ell}\right |,$$ where $\mathcal{A}$ is all possible ways to get $k$ different subsets $\{A_i\}$ of $[n]$ and $\mathcal{A}_{\ell}$ is the number of ways to get $k$ different subsets of $[n]$ such that $\ell$ is in non of them. Show that
$|\mathcal{A}|=\binom{2^m}{k},$ and that if $X\subseteq [m],$ then  $$\left |\bigcap _{x\in X}\mathcal{A}_x\right |=\binom{2^{n-|X|}}{k}.$$
Notice that $$\mathcal{A}\setminus \bigcup _{\ell=1}^m\mathcal{A}_{\ell}$$ is exactly the objects you want. You eliminate every time their union was not exactly $[n].$
