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.
I observe that each term with $i$ fixed approaches a nice limit. We have
$$ 2^i \frac{n(n-1)(n-2)\cdots(n-i+1)}{i!} \frac1{(2n-i+1)(2n-i+2)\cdots(2n)} $$
or
$$ \frac1{i!} \frac{2n}{2n} \frac{2(n-1)}{2n-1} \cdots \frac{2(n-i+1)}{(2n-i+1)} \sim \frac 1{i!} $$
This gives you the series, assuming the limits (defining terms with $i>n$ to be zero) may be safely exchanged,
$$\lim_{n\to\infty} \sum_{i=0}^\infty [\cdots] = \sum_{i=0}^\infty \lim_{n\to\infty} [\cdots] = \sum_{i=0}^\infty (-1)^i \frac1 {i!} \equiv e^{-1}$$
Justifying the limit interchange I haven't thought about but I suspect this can be shown to be fine without too much effort... Edit: You can probably use the Weierstrass M-test.
Best Answer
For this particular problem there is an easy way to count (but right now I don't see how to generalise this to more than $4$ couples). One can first seat the women in $4$ alternating seats. Assuming you mean to identify rotationally symmetric arrangements this can be done in $3!=6$ ways: the first women serves as reference and her seat can be numbered 0, and seating the three other women is given by a bijection to the seats $2,4,6$. (If you also want to identify reflection symmetry, divide by $2$.)
Now to seat the men, there are two options for the husband of the lady in seat $0$, namely seats $3$ and $5$. But when this is done, the arrangement is fixed. Supposing he took seat $3$, then this seat is no longer available for the husband of the lady in seat $6$, who then must go to seat $1$; then the husband of the lady in seat $4$ must go to seat $7$, and the remaining husband to seat $5$. In case the first husband took seat $5$, the situation is similarly fixed, reasoning in the opposite direction. So in all there are $6\times 2=12$ solutions.
Added I finally found out that for $n$ couples the number is given (up to a factor $(n-1)!$ for seating the women first) by A000197 in OEIS. Presumably you can find useful things in the comments and formulas there.