If you want the expected value, one answer is $n E[S_{(m)}]$, where $S_{(m)}$ is the $m$th order statistic of a gamma$(k,1)$ random variable. While this expression may not have a simple closed form, you may be able to get a decent-sized approximate answer from the literature on moments of order statistics.
Here's the argument: Take a Poisson process $P$ with rate 1 and interarrival times $Z_1, Z_2, \ldots$. Let each event in the process $P$ have probability of $1/n$ of being the first kind of coupon, probability $1/n$ of being the second kind of coupon, and so forth. By the decomposition property of Poisson processes, we can then model the arrival of coupon type $i$ as a Poisson process $P_i$ with rate $1/n$, and the $P_i$'s are independent. Denote the time until process $P_i$ obtains $k$ coupons by $T_i$. Then $T_{i}$ has a gamma$(k,1/n)$ distribution. The waiting time until $m$ processes have obtained $k$ coupons is the $m$th order statistic $T_{(m)}$ of the iid random variables $T_1, T_2, \ldots, T_n$. Let $N_m$ denote the total number of events in the processes at time $T_{(m)}$. Thus $N_m$ is the random variable the OP is interested in. We have
$$T_{(m)} = \sum_{r=1}^{N_m} Z_r.$$
Since $N_m$ and the $Z_r$'s are independent, and the $Z_r$ are iid exponential(1), we have
$$E[T_{(m)}] = E\left[E\left[\sum_{r=1}^{N_m} Z_r \bigg| N_m \right] \right] = E\left[\sum_{r=1}^{N_m} E[Z_r] \right] = E\left[N_m \right].$$
By scaling properties of the gamma distribution, $T_i = n S_i$, where $S_i$ has a gamma$(k,1)$ distribution. Thus $T_{(m)} = n S_{(m)}$, and so $E\left[N_m \right] = n E[S_{(m)}]$.
For more on this idea, see Lars Holt's paper "On the birthday, collectors', occupancy, and other classical urn problems," International Statistical Review 54(1) (1986), 15-27.