By the condition, it is known that $1$, $2$, $4$, $m$, $m^2$ (allowing for repetition) are all the elements of $C$.
Note that when $m$ is real, $1+2+4+m+m^2 > 6$, $1+2+4+m^2 > 6$, so it can only be $C = \{1, 2, 4, m\}$, and $1+2+4+m = 6$. Therefore, $m = -1$, and is tested to be consistent with the question. At this point the product of all the elements of $C$ is $1 \times 2 \times 4 \times (-1) = -8$.