For example, given $n=2$, we only need 2 points, (for example (0,0) and (1,1)), such that any other corner can be reached by only traveling 1 edge at max. For $n=3$ this is true as well that only 2 points are needed.
However this is not the case for $n=4$, is there an easily generalized formula for the number of points required to achieve this?
It's easy to give a formula for a lower bound: each selected point accounts for itself and $n$ neighbors, so at least $\big\lceil \frac{2^n}{n+1}\big\rceil$ points are required. For $n=4$, this gives $4$, which is correct. (The points could be $(0,0,0,0), (0,1,1,1), (1,0,0,0)$ and $(1,1,1,1)$, for example.)
Unfortunately, this lower bound cannot always be achieved. For $n=5$, the lower bound gives $6$, but $7$ points are actually needed. The values for $n \le 9$ are given in OEIS sequence A000983, but the sequence is labelled "hard" which suggests that not only is there no exact formula known, but also no efficient method for computing further values. Even for $n=10$, the answer is not known.
A significant exception (duly noted on the sequence's page) is that when $n$ is of the form $2^k - 1$ the lower bound can be achieved. In this case, the minimal covering of the $n$-hypercube corresponds to a Hamming code.