8. Each permutation of $1,2, \cdots, 8$ is considered as an eight-digit number. Then the number of eight-digit numbers that are multiples of 11 is $\qquad$.
8.4 608 .

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

and $S(A) 、 S(B)$ represent their digit sums. Suppose
$$
S(A) \geqslant S(B) \text {. }
$$

Then $S(A)+S(B)=36$.
Thus, $S(A) 、 S(B)$ 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 where the four numbers are in odd or even positions, there are two cases. Therefore, the total number of eight-digit numbers that satisfy the condition is
$$
4 \times 2 \times 4!\times 4!=4608 \text { (numbers). }
$$