Suppose we have an fair $m$-sided die, and we roll it $n$ times. What is the expected frequency $E(n, m)$ of the most frequently rolled face?
If we fix $n$ we can calculate $E(n,m)$ like so. Let $\Pi(n)$ be the set of partitions of $n$, where each partition is a weakly descending sequence of nonnegative integers, like $(5,2,2,1,1,0...)$. For each partition $\pi$, we calculate the probability that the frequency distribution equals $\pi$, then we multiply that by $\pi_1$ to get the contribution to $E$. Summing over $\Pi(n)$ gives us $E$.
I have computed the following (using Mathematica):
$\begin{align*} E(1,m) &= 1\\ E(2,m) &= (m+1)/m\\ E(3,m) &= (m^2+3m-1)/m^2\\ E(4,m) &= (m^3+6m^2-7m+4)/m^3\\ E(5,m) &= (m^4+10m^3-25m^2+40m-21)/m^4\\ E(6,m) &= (m^5+15m^4-65m^3+195m^2-266m+126)/m^5\\ E(7,m) &= (m^6+21m^5-140m^4+665m^3-1631m^2+1911m-820)/m^6\\ E(8,m) &= (m^7+28m^6-266m^5+1820m^4-6881m^3+14140m^2-14554m+5720)/m^7 \end{align*}$
The probability that the maximal face has the frequency $k$, which appears $t$ times, is $$ \begin{align*} p_{k,t} &= m^{-n} \binom{m}{t} \left[(1+\cdots+x^{k-1})^{m-t}\right]|_{x^{n-kt}} \\ &= m^{-n} \binom{m}{t} \left[ \frac{(1-x^k)^{m-t}}{(1-x)^{m-t}} \right]|_{x^{n-kt}} \\ &= m^{-n} \binom{m}{t} \sum_{s=0}^{m-t} (-1)^s \binom{m-t}{s} \binom{n-kt-(s-1)(m-t)-1}{m-t-1}. \end{align*} $$ Here the binomial coefficients should be understood as null if "out of range". Given $p_{k,t}$, the expectation you are after is $\sum_{k,t>0} kp_{k,t}$. I'm not sure there is any more succinct way to write this expression. Compared to your formula, though, this expression should be more efficient to compute. However, this expression doesn't make it immediately clear why the result should be polynomial in $m$, given $n$.