Consider a Turing machine M1 of size X that, given an input value n, calculates a function f(n) such that O(f(n)) > O(BB(n)) and halts.
Consider a Turing machine M2 of size Y that calculates a number A and halts, where X + Y < A.
Combine the two machines, using the number calculated by M2 as the input for M1. The resulting Turing machine of size X + Y calculates a value f(A) such that f(A) > BB(A) > BB(X + Y) and then halts. This contradicts the definition of BB(X + Y), therefore no such computable function f can exist.
It's more of a proof sketch, really--there's a few unexamined assumptions in there. It was the best I could do off the top of my head and typing on a smart phone, but I'm reasonably confident that the overall structure is valid.
I was actually kinda hoping someone who knows more than I do would chime in and clarify things...
Well, here's another approach. We can solve the halting problem for any finite TM (http://portal.acm.org/citation.cfm?id=1052796.1052798), so let's imagine we're looking at a possible BB, n. All we have to do is run all the possible programs, and see whether of the halting ones, one is bigger than n.
Consider a Turing machine M2 of size Y that calculates a number A and halts, where X + Y < A.
Combine the two machines, using the number calculated by M2 as the input for M1. The resulting Turing machine of size X + Y calculates a value f(A) such that f(A) > BB(A) > BB(X + Y) and then halts. This contradicts the definition of BB(X + Y), therefore no such computable function f can exist.