> But imagine that we could compute BB(n) for any n, even if we need to use a different algorithm for each n.

This is itself uncomputable; therefore

> Then, we could build the mapping n -> BB(n).

This mapping is not computable.

Edit as I'm rate-limited:

What I mean is, informally, the incomputability of the algorithm to select the correct algorithm to compute BB(n) is sufficient to make that mapping uncomputable.

However, that does not mean that an algorithm to compute BB(n) for a specific n does not exist. (Trivially, it does; "output BB(n)".)