computational complexity - BPP being equal to #P under Oracle

First, let's be slightly pedantic and not make statements like P = #P, which cannot possibly be true just because P is a set of decision problems and #P is not. To get a decision version of #P, one can use PP, or something like P^#P.

About your question, BPP^NP is contained in P^PP and P^#P by Toda's theorem. On the other hand, if P^#P were contained in BPP^NP, it would imply that PH is contained in BPP^NP, which would collapse the polynomial hierarchy to the third (or second?) level, which is considered unlikely.

In conclusion, P^#P is considered to be more powerful than NP, BPP, BPP^NP and even NP^NPNP.