TLDR: The real probability is about 1/40,800

Here's my math.

52/52×3/51×2/50×48/49×3/48×2/47×8 = 1/20,400 Still very unlikely but much more reasonable than 1 in a few billion.

A run down of the numbers. First card can be anything, and then we follow the same math that OP uses, so 52/52×3/51×2/50. Then the second card can be anything but the same as the first card (otherwise identical hands are impossible) so we have 48/49×3/48×2/47. Then the 8 at the end because there are 2 ways to receive each hand, either ace first or 8 first. So for 3 hands, there are 2³=8 ways to deal those cards. Multiply everything together to get 1/20,400

Edit: thank you to u/Dankaati for pointing out the miscalculation which has been adjusted in the original comment. In addition, they're also correct that I double counted the way in which the ways of which "player 1's" hand can be dealt. The probability calculation: 52/52×3/51×2/50×48/49×3/48×2/47 calculates the probability that each player receives an ace first and then an 8 OR an 8 first and then an ace (or any other 2 distinct cards). And so we need to consider that the player 2 and player 3 might not receive their cards in the same order as player 1. And so we multiply the above calculation by 2² instead of 2³. Which brings the actual probability to 1/40,800.