Yes. A valid question for a specific n would be whether you can prove the value of BB(n). If you don't care about provability, you can indeed just produce a number that happens to be the right one.

So as you noticed, it only makes sense to talk about whether a function is computable, we can't meaningfully talk of computable numbers.

Yes exactly, imagine a function HH(n) that returns 0 if the Turing machine represented by the integer n halts, and 1 if it doesn't.

Then HH the function itself is not computable, but the numbers 0 and 1, which are the only two outputs of HH are computable.

Integers themselves are always computable, even if they are the output of functions that are themselves uncomputable.

The main thing to make it clear is that BB(n) for a specific n isn't a function - it's a number. Just like Mult(10,4) isn't a function, it's a number (40).

So a specific BB(n) is just a number and is computable.