Clearly, the product of the eight products is exactly $(8!)^{24}$ so by AM-GM the sum is at least $8 \cdot (8!)^3$. On the other hand, equality can certainly be achieved by starting with $1,2,3,4,5,6,7,8$ in the first row, then $2,3,4,5,6,7,8,1$ in the second one etc. (shifting by $1$ each row) so that the product in each column is just $(8!)^3$.