First, judging from your work, you misstated the question: it appears that you meant to ask for the probability that no husband sits next to his wife. It also appears that the seats are arranged in a row, not a circle. I will make these assumptions.
Your first three factors, $8\cdot6\cdot5$, in the numerator are fine. After that, though, you have to split the calculation into cases. Suppose that the first three people, in order, are A, B, and C. If A and C are a couple, then any of the remaining $5$ people can sit in the fourth seat. Your calculation, with a factor of $4$, is correct only if C’s spouse has not already been seated, i.e., C and A are not a couple. And the cases just proliferate after that, which is why you’re better off working with the complement.
I cannot think of anything pleasant. A natural approach is through Inclusion/Exclusion.
There are $8!$ arrangements. If we can count the bad arrangements, in which at least one couple is together, then the rest is easy.
Call the couples A, B, C, D and let $X$ be the wife in couple X, and $x$ the husband. It is not hard to count the arrangements in which $a$ is next to $A$, and similarly for the other $3$ couples.
If we add these $4$ numbers, we will have double-counted, in particular, the arrangements in which couple A and couple B are both together. So we need to subtract $\binom{4}{2}$ times the number of arrangements in which couple A and couple B are together.
But we have subtracted too much. So we add back $\binom{4}{1}$ times the number of arrangements in which couples A, B, and C are together.
But we have added back too much, so we must subtract the number of ways for all the couples to be next to each other.
Instead of counting, we can apply Inclusion/Exclusion directly to probabilities. It is marginally easier.
Best Answer
Assuming you seat the 8 individuals at random, one way of doing this is to use the inclusion exclusion principle and turn a number of couples into virtual individuals so they must sit next to each other to get $$1-\frac{ 2^1{4 \choose 1} 7! - 2^2{4 \choose 2} 6! + 2^3{4 \choose 3} 5! - 2^4{4 \choose 4} 4!}{8!}$$