I ran a computer program, brute-forcing through all possibilities. Assuming my program is correct, the best choice for the first player is
6, giving them a probability of 31/96 to be a winner.
After this, play continues with
11 for the second player, giving them a 30/96 probability of being a winner.
8, 9, or 10 for the third player, each giving probability 26/96.
3 for the fourth player, with a probability between 11/96 and 13/96 depending on the third player's choice.
Fifth player cannot win, so chooses any remaining number at random.
Note that the probabilities add to more than 1 since there can be more than one winner.
Obviously, if the last player has a winning move, they will choose it. For any of the first players to win, they need to choose numbers such that player five's winning move is already taken.
This usually forces player four's move. Player four must choose the number that player five would want to play. So it must be such that if five played the same then it is about half the average. A little algebra shows that it should be (a+b+c)/8, where a,b,c are the moves of the first three players. In the optimal playing sequence above that is indeed the case.
I have only a rough explanation for the first three plays. The fifth player's random choice can shift the half-average by about 10. So ideally the first three moves should cover most of such a range about equally, forcing the fourth player to accept a smaller cut at the low end of the range.
My program previously counted shared wins differently, by spreading the probability amongst them. This is equivalent to one of the winning players being chosen at random to be the sole winner of the round. If that variation of the game is used, then play is slightly different.
Player one plays 11 first, then player two chooses 6 or 7, player three has a few more choices in the 5-10 range, and then player four still chooses 3.