The number of good states with $k$ lights on in each row is ${\binom nk}^2$; summing over $k$, there are $\sum_{k=0}^n{\binom nk}^2 = \sum_{k=0}^n \binom nk \binom n{n-k}$ good states. This is the $x^n$ coefficient of $(1+x)^n(1+x)^n = (1+x)^{2n}$, i.e. $\binom{2n}n$. Hence the desired value is $$\frac{\binom{2n}n}{2^{2n}} = \frac{(3\cdot 5 \cdots (2n-1))(2\cdot 4 \cdots 2n)}{2^{2n}n!(1\cdot 2 \cdots n)} = \frac{(3\cdot 5 \cdots (2n-1))(2^n)}{2^{2n}n!} = \frac{3\cdot 5 \cdots (2n-1)}{2^{n}n!}.$$

Consider the operation of switching every light in the second row from on to off, and from off to on. This is a bijection between the set of all good states, and the set of states with $n$ lights on and $n$ lights off. Hence, number of good states is $\binom{2n}n$. Since total number of states is $2^{2n}$, the conclusion follows.

Vandermonde's identity trivializes the problem