Having thought about the matter somewhat, there's an cute half-way solution to the problem with fewer padlocks (8) and less chain shenaniganing. Label the pirates A to E, and label padlocks with the pirates who have the relevant keys (so ABC can be opened by pirates A, B or C).
Set up padlocks ABC, ABD, ACD and BCD locked round the clasp. Set up four more padlocks AE, BE, CE and DE, attached to the chest and pairwise connected by six lengths of chain through the clasp, such that the removal of the padlock at either end of a chain allows the chain to be drawn out of the clasp.
Given E and one other pirate (WLOG A), the chest is unopenable because BCD will remain locked. Given any additional pirate, the chest is openable. Given two non-E pirates (WLOG A and B), the chest is unopenable because the the chain linking CE and DE remains locked. Given any additional pirate, the chest is openable.
I think that this is minimal for direct linkages, but am not quite convinced.