The answer is 57%.
The problem is that the following reasoning is false: the two aces will be distributed uniformly and independently amonst the other three players. The probability that my partner receives at least one is 1 - (2/3)(2/3) = 55%. Although close, the estimate falls short of the truth due to the unfounded assumption of independence.
Consider my six cards removed from the deck, and the remaining 18 cards collected into triples, each triple is dealt on a single round. If the two aces are in distinct triples, then the distribution of aces is independent. If the two aces are in a single triple, then the distribution of aces is highly dependent. In fact, in this case, where three cards are dealt and two are aces, it is 66% certain that my partner will be dealt an ace.
In this second case, the aces push themselves away from each other. If you have one ace, you certainly must not have the other. The true result is a affine linear combination of the two cases. So this pushing away is still a true effect, seen diluted by the probabilistic weight of the competing cases.