Define the function $F^*(n)=\underset{p_1\dots p_k\leq n}{\sum_{p_, \ldots, p_k}}\frac{1}{p_1\dots p_k}$. Then we have for each fixed $k$ the asymptotics $F_k^*(n)\sim(\log\log n)^k$. To see this note the obvious upper and lower bounds
$$
\left(\sum_{p\leq n^{1/k}}\frac{1}{p}\right)^k\leq F_k(n)\leq \left(\sum_{p\leq n}\frac{1}{p}\right)^k,
$$
and use the asymptotics $\sum_{p\leq n}\frac{1}{p}\sim\log\log n$.

To deduce an asymptotics for $F_k(n)$, note first that summands in $F_k^*$ corresponding to tuples which are not pairwise different are negligible, in fact, using a variation of the upper bound above we see that this contribution is $\mathcal{O}((\log\log n)^{k-1})$. If the primes are pairwise different, then the corresponding summand occurs $k!$-times in $F_k^*$ and only once in $F_k$, thus $F_k(n)\sim\frac{1}{k!}(\log\log n)^k$, confirming your computation.

As $k$ grows, we use the bound $\sum_{p\leq n}\frac{1}{p}=\log\log n +\mathcal{O}(1)$, thus $F_k^*(n)=(\log\log n)^k+\mathcal{O}(k(\log\log n)^{k-1})$. Deleting tuples which contain the same prime multiple times gets more complicated. A crude estimate would be $B_k(\log\log n)^{k-1}<k^k(\log\log k)^{k-1}$, where $B_k$ denotes Bell numbers. In this way you get a uniform bound almost up to $k=\log\log\log n$. If you need better estimates, you would have to consider partitions more carefully, which is doable, but some work.