How many $3 \times 3$ integer matrices are orthogonal?

Dot product of columns can be used (it must be $0$) The first column of matrix $A$ has 6=2*3 possibilities for location 1 or -1, after location 1 or -1 the rest of entries must be zeros and they give possibilities for the second column - we have here only 2*2 = 4 possible choices for location 1 or -1, and the third column stays with only 2.
6*4*2=48.

But rotation matrices in right handed frame (if you would want only them in the future) we have only 24, because the third column is always calculated as the cross product of 1 and 2 column.

Note that each column of $A$ must be an integer vector of unit length which means that each column is of the form $\pm e_i$ for some $1 \leq i \leq 3$ (where $e_i$ are the standard basis vectors). Thus, we need to pick a permutation of the $e_i$'s to put as columns and then, for each column independently, decide whether it gets a plus or a minus sign. This results in a total of $3! \cdot 2^3 = 6 \cdot 8 = 48$ options for $A$.