The probability of the complementary event, that no face appears $m$ times or more, is
$$
\frac{n!}{k^n}[x^n]\left(1+\frac{x^1}{1!}+\frac{x^2}{2!}+\dots+\frac{x^{m-1}}{(m-1)!}\right)^k\tag{$*$}
$$
Here, $[x^n]f(x)$ means the coefficient of $x^n$ in the polynomial $f(x)$ (more generally, $f$ can be a power series). To see this, note that if you brute force expand out $\left(1+\frac{x^1}{1!}+\frac{x^2}{2!}+\dots+\frac{x^{m-1}}{(m-1)!}\right)^k$ and collect all powers of $x^n$, the result is
$$
\frac1{k^n}\sum_{a_1+a_2+\dots+a_k=n,\\0\le a_i\le m-1}\frac{n!}{a_1!a_2!\cdots a_k!}
$$
The summation ranges over all tuples $(a_1,\dots,a_k)$ of nonnegative integers between $0$ and $m-1$ whose sum is $n$. For each $i\in \{1,\dots,k\}$, the variable $a_i$ indicates choosing the summand $x^{a_i}/a_i!$ in the $i^{th}$ copy of $(1+x^1/1!+\dots+x^{m-1}/(m-1)!)$. But this summation is obviously a brute-force way to count the number the number of sequences of length $n$, whose entries are die faces, such that each face appears at most $m-1$ times.
This also follows from the the theory of exponential generating functions, which is explained in chapter 3 of generatingfunctionology, available for free online.
Finally, the answer to your question is one minus the expression $(*)$.