(Un)computability of a restricted Halting Problem
Fix a value of $n$. For $b \in \{0,1\}^n$, consider the following algorithm $A_b$:

If $x \leq n$ then output $b_x$, otherwise output $0$.

Clearly one of the $2^n$ algorithms of the form $A_b$ computes your function $h$, hence $h$ is computable.