Thanks for the answers so far to this question. I think I have now solved the easier part of the question, using an inclusion-exclusion argument (as you suggest Ross) to count the number of possible combinations of hands where there is a card common to all players.
Suppose there are 3 players ($n=3$) and each draws 2 cards ($m=2$). The total number of possible arrangements of different hands is then $\binom{52}{2}^3$. Of these arrangements, the number where every player's hand includes some named card (say, ace of spades) is $\binom{51}{1}^3$. Summing over all possible cards gives $52 \times \binom{51}{1}^3$ but this will double count the $\binom{52}{2}$ arrangements where every player holds the same two cards. So the total number of arrangements where at least one card is shared is $52 \times \binom{51}{1}^3 - \binom{52}{2}$, and the probability of such an arrangement is:
$$\frac{52 \times \binom{51}{1}^3 - \binom{52}{2}}{\binom{52}{2}^3}$$
Generalising this argument to higher m and n gives the general formula for the probability:
$$\frac{\sum_{i=1}^{m}(-1)^{(i+1)}\binom{52}{i}\Big(\binom{52-i}{m-i}^n\Big)}{\binom{52}{m}^n}$$
I've done some quick calculations in Excel to check the case $m=13$ and agree with the numbers provided by Robert.
Moving on to the harder part of the question, I have found the probability that there are k (or fewer) cards that "cover" all players for the special case where $m=1$ (each player picks one card). Counting the ways of needing exactly j cards to cover all individuals, using an inclusion-exclusion argument, and summing j from 1 to k gives:
$$\frac{\sum_{j=1}^{k}\binom{52}{j}\sum_{i=0}^{j-1}(-1)^i\binom{j}{j-i}(j-i)^n}{52^n}$$
However, as yet I've got no further with generalising for $m>1$...