We check your count of the number of ways at least one couple are in adjoining chairs. As you did, we use Inclusion/Exclusion. Call the couples A, B, and C.
The number of ways couple A are next to each other is $(2)(4!)$. For we may assume the fatter of the two members sits in a specific chair. Then the partner has $2$ choices, and for every such choice the rest can arrange themselves in $4!$ ways.
Add together the number of ways for couple A to be together, couple B to be together, couple C to be together. We get $144$.
But we have overcounted the ways in which for example couple A and B are both together. Again we can assume the fatter of the members of A sits in a specific chair. Then there are $2$ choices for where the other member sits, and then $(3)(2)$ ways for couple B to be together, leaving $2$ choices for the others, a total of $24$. Summing over all ways to choose $2$ couples, we get $72$. So our corrected estimate is $144-72$.
But the $144$ counted three times the arrangements in which all the couples are together, and so did the $72$, so we must add back the $16$ ways in which they are all together. The final count is $144-72+16$.
Note that this $88$ counts the number of ways at least one couple are together. It is naturally a much larger number than the $16$ ways in which all couples are together.
Here's a different approach. There are four surnames A B C and D. If you draw a dividing line across the middle of the table, then on one side you need to arrange the letters A B C D which can be done in $4!$ ways.
For each of these surnames, there is a husband or a wife who occupies this position, with their spouse automatically sat in the place opposite. We can choose whether each surname is represented by husband or wife in $2^4$ ways.
Finally, for each arrangement, we can ask everyone to move one place to the left, but we still have the same arrangement. Likewise we can rotate each arrangement into eight new positions but each arrangement is still the same as the original. We therefore divide by eight.
Putting this all together we have a total of $$\frac{24\times2^4}{8}=48$$ arrangements.
Best Answer
This is a variant of the Menage Problem and a solution of that can be found in one of Bogart's articles(no surprise there) itself: http://www.math.dartmouth.edu/~doyle/docs/menage/menage/menage.html
(See the solution to the Relaxed Menage Problem).