Order of the smallest group containing all groups of order $n$ as subgroups.

This is not known. In fact we don't even know how many groups of order $n$ exist for most $n$. A related open question asks what the smallest integer $s$ such that all groups of order $p^r$ embed in some group of order $p^s$.

However, I have reason to doubt that the number will be $m!$ for $m=p^e$. (This is not a proof, by the way, only a heuristic argument.) As in my link the growth rate of $p$-group isomorphism classes is $$p^{\left(\frac{2}{27}+O(n^{-1/3})\right)n^3}.$$ Multiply this by $p^n$ to count the number of elements in each subgroup (which assumes generously they will be embedded without overlap, including the identity) and we get $$p^{\left(\frac{2}{27}+O(n^{-1/3})\right)n^3+n}.$$ Comparing this to Stirling's approximation $$p^n!\sim \sqrt{2\pi p^n}\left(\frac{p^n}{e}\right)^{p^n}$$ you can see that the $p^n!$ will quickly become much too large to be the smallest group containing these isomorphism classes even if we were to insist that the groups were embedded completely separately.

In fact, we can use Stirling's approximation along with the upper bound in Gallagher (1967) to show that the number will almost never be $m!$ for large $m$. $$\lim_{m\to\infty}\frac{m^{cm^{2/3}\operatorname{log}(m)+1}}{\sqrt{2\pi m}\left(\frac{m}{e}\right)^m}=\lim_{m\to\infty}\frac{e^m m^{\frac{1}{2}-m+cm^{2/3} \operatorname{log}(m)}}{\sqrt{2 \pi }}= 0$$

So, I suspect that in general the number will be much lower than $m!$.

For a counterexample outside $n=p$ prime (where obviously $m=p$, since every order $p$ group is isomorphic to $\mathbb{Z}/p\mathbb{Z}$), consider $n=15$.

Then every order $15$ group is Abelian and is actually isomorphic to $\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}=\mathbb{Z}/15\mathbb{Z}$.

So $m=15$ is the answer in this case.

Edit: of course, each time there is only one group of order $n$, the answer is $m=n$. See here for the beginning of a list of these cases.