How could you "solve" the halting problem if, hypothetically, the busy beaver numbers were "small"? I read that if BB(n) did not grow faster than all computable sequences of integers, you could solve the halting problem and contradict Turing's theorem.

I'm trying to figure out how you could specifically do this. One obvious possibility is to build an n-state Turing machine that looks for a solution to an undecidable problem, then simply waiting for it to pass BB(n) transitions. This should be feasible since BB(n), in this scenario, is small, and surpassing it would prove that the machine loops forever.

But this isn't a satisfactory answer because surely decidability isn't about how feasible a procedure is in practice, but whether it's decidable in theory - what if computers were powerful enough to be able to implement this in our current universe where BB(n) is very large? Shouldn't we just assume that the machine is independent of number theory instead?
To follow the strategy you outline, we need to be confident a priori that we have an upper bound on $BB(n)$. More snappily, if $f$ is any function with $f(n)\ge BB(n)$, then from $f$ we can compute the halting problem: given an $n$-state machine, run it for $f(n)$-many steps.

Since the halting problem isn't computable, this means that no such $f$ is computable; this is a "global" phenomenon. No particular value of $BB$ is too large to be used, but rather the growth rate of $BB(n)$ prevents us from having a whole sequence of upper bounds, one for each $n$. As a low-level analogy:


  There is no polynomial function which is always above $exp(n)=2^n$. Of course, this isn't because individual values of $exp$ are too big, but rather because of the overall behavior of the function across all inputs
  .