Look, by Euler's theorem the number of partitions $p_{dist}(n)=p_{odd}(n)$. Since the number of parts of a random odd partition (i.e. into odd part sizes) is about $O(\sqrt{n}\log n)$, removing 1 from each part gives an even partition of roughly the same size. This gives $$p_{odd}(n) \approx p_{even}(n) = p(n/2), $$ when defined appropriately. This shows that you really don't need a separate new asymptotic formula for $p_{dist}(n)$ if rough approximation is ok. While informal, this argument can be made completely formal, and has been done a few times, I think.

Look, by Euler's theorem the number of partitions $p_{dist}(n)=p_{odd}(n)$. Since the number of parts of a random odd partition (i.e. into odd part sizes) is about $O(\sqrt{n}\log n)$, removing 1 from each part gives an even partition of roughly the same size. This gives $$p_{odd}(n) \approx p_{even}(n) = p(n/2), $$ when defined appropriately. This shows that you really don't need a separate new asymptotic formula for $p_{dist}(n)$ if rough approximation is ok. While informal, this argument can be made completely formal, and has been done a few times, I think.

UPDATE: I came across Hua Luogeng's paper "On the number of partitions of a number into unequal parts" (1942), which gives an analogue of Rademacher's formula for $p(n)$.