I don't think any nice solutions to this can exist. However I offer the following possibility:
(6 4 2 5) (1 3 7 8)
(3 5 4 1) (2 6 7 8)
(3 6 1 2) (4 5 7 8)
(1 2 5 7) (4 3 6)
(5 6 7 3) (4 1 2)
(7 3 2 4) (5 1 6)
where player 8 is the one who drops out after 3 days. Perhaps there might be better possibilities, but in order to have everyone play each other at least once, and no more than three times, then I end up with 7 and 8 playing together in all of the 1st 3 rounds.