This is my second try, but I think is a MUCH better argument than what I had before. So I deleted the more complicated answer I tried to give earlier.
The total # of possible deals to the two players is: $$\binom{52}{2}\binom{50}{2}=1,624,350$$
(i.e., choose two cards for the first player and then two for the second.)
Then, the # of ways to deal a blackjack to both players would be: $$\binom{4}{1}\binom{16}{1}\binom{3}{1}\binom{15}{1}=2880$$
(i.e, choose which ace to give to the first player, then which 10,J,Q,K for the first player, then which ace for the second player, and then which 10,J,Q,K for the second player.)
Also, the # of ways that exactly one player gets a blackjack would be: $$\binom{2}{1}\binom{4}{1}\binom{16}{1}\times(\binom{3}{2}+\binom{3}{1}\binom{32}{1}+\binom{47}{2})=151,040$$
(i.e., choose which player to give the blackjack, then choose which ace to give them, and then which 10,J,Q,K to give them. Then for the other player we either give them two aces from the 3 that are left, give them one ace from the 3 that are left and 1 of the 32 2-9 cards, or give them two non-aces from the 47 non-aces that are left.)
So, the probability of at least one player getting a blackjack is: $$\frac{2880+151,040}{1,624,350}$$
Hence the probability of neither player getting a blackjack is: $$1- \frac{2880+151,040}{1,624,350} \approx 90.5\%$$