Initial thoughts; will come back to this:
Start with a small case, 3 varieties. Say there are 5 samples. How could we have an empty box? Either one or two empty boxes:
P(one box empty) = 3*(2/3)^5 = 32/243 [3 choices empty box, 2/3 chance in some other box]
P(two boxes empty) = C(3,2)*(1/3)^5 = 1/243 [3 choices of two empty boxes, 1/3 chance to land in other box]
"One box empty" includes the possibility of two boxes empty, so by inclusion / exclusion, P(nothing empty) = 1 - 32/243 + 1/243 = 214/243.
To generalize, you want some sort of asymptotic for the PIE computation, probably involving Stirling's approximation for factorials.