All numbers divisible by 8 has the property that the last 3 digits are also divisible by 8. But note that since the 4-digit number is a palindrome, we are actually counting the number of 3-digit multiples of 8 that has the same first two digits, and does not end with 0. We can then make put the last digit in front to get back the 4-digit palindrome we wanted. The other case is that the number ends with 00(digit), so we need to take that into account as well, but this case is simple because the only possibility is 8008.

For the first case, we first count the number of 3-digit multiples of 8 that has the same first two digits. They are 112, 224, 336, 440, 448, 552, 664, 776, 888, and 992. So in total 10, but 440 ends with 0 so we exclude that case. Hence, we have only 9 cases here, which gives the palindromes 2112, 4224, 6336, 8448, 2552, 4664, 6776, 8888, 2992 respectively.

Including 8008, the answer is 10.