Okay this is a bit tricky for me. Jemidiah is definitely one guy who knowss how to do this. Hopefully he'll chime in, but here's my attempt:

okay so the way I see it is first we need to choose rank, then suit then the number of ways to get dealt the 3 of a kind and then finally the rest of the hand.

There are 5 ranks, 4 suits, and 4 of each suit per ranks so:

number of combinations of getting AT LEAST 3 (you may get another triple or more of another rank) of each either A/K/Q/J/10 is:

first triple: (5,1) (4,1) (4,3)
second triple given first: (1) (3,1) (4,3)
third triple given others: (1) (2,1) (4,3)
4th triple given others: (1) (1,1) (4,3)

There are a total of 80 cards in the deck but we remove the selected rank leaving 80 - 16 = 64.

Number of ways to deal 8 cards from 64: (64, 8)

this gives a total of: 4^4 * 5! * (64,8) = 135971800104960

total number of hands: (80, 20) = 3535316142212174320

probability = 3.846e-5

Does that jive with your experimental results? Do you allow the possibility of other triples in the hand?