8. Consider each permutation of $1,2, \cdots, 8$ as an eight-digit number. Then the number of eight-digit numbers that are multiples of 11 is $\qquad$
8. 4608 .

For each such eight-digit number $\overline{a_{1} a_{2} \cdots a_{8}}$, let
$$
A=\left\{a_{1}, a_{3}, a_{5}, a_{7}\right\}, B=\left\{a_{2}, a_{4}, a_{6}, a_{8}\right\} .
$$

Let $S(A)$ and $S(B)$ denote their digit sums, and assume $S(A) \geqslant S(B)$.
Then $S(A)+S(B)=36$.
Thus, $S(A)$ and $S(B)$ have the same parity, and $S(A)+S(B)$ and $S(A)-S(B)$ have the same parity.

Therefore, $S(A)-S(B)$ is an even number and a multiple of 11. If $S(A)-S(B)=22$, then $S(B)=7$, which is impossible (since the sum of the smallest four numbers is at least 10).
Thus, $S(A)-S(B)=0$, i.e.,
$S(A)=S(B)=18$.
Hence, $S(A)=S(B)$ has only four cases:
1278, 3456; 1368, 2457;
1467, 2358; 2367, 1458.
In each group, the four numbers can be fully permuted, resulting in $4! \times 4!$ eight-digit numbers.

Considering the two scenarios for the four numbers in odd or even positions, the total number of eight-digit numbers that satisfy the condition is
$$
4 \times 2 \times 4! \times 4! = 4608 \text{ (numbers). }
$$