The answer is $0$ if $n = 1$, and is $$\sum_{k=0}^{n} (-1)^k \binom{n}{k} 2^k (2n-k-1)!$$ if $n \ge 2$, or any equivalent expression.

The result is obvious for $n = 1$. Consider $n \ge 2$. For each $k = 0, 1, \dots, n$, let $N_k$ be the number of ways to arrange the $2n$ people such that $A_i, B_i$ are next to each other for $i = 1, 2, \dots, k$ (and probably for some more pairs).

For each $k$, we need to arrange $k$ pairs and the remaining $2n-2k$ people. There are $(k+2n-2k-1)! = (2n-k-1)!$ ways to do so. Within each pair, there are $2$ ways to arrange the two people. Therefore, we have
$$N_k = 2^k (2n - k - 1)! .$$
Now, by the inclusion-exclusion principle, since there are $\binom{n}{k}$ ways to choose $k$ pairs, the number of ways to arrange the $2n$ people such that no pair $A_i, B_i$ sits next to each other is
$$\sum_{k=0}^{n} (-1)^k \binom{n}{k} N_k = \sum_{k=0}^{n} (-1)^k \binom{n}{k} 2^k (2n-k-1)! .$$