Square-root cancellation for the signs of Latin squares

Let $L(n)$ be the number of Latin squares of order $n$, and let $L^{\textrm{even}}(n)$ and $L^{\textrm{odd}}(n)$ be the number of even and odd such squares, so that $L(n) = L^{\textrm{even}}(n) + L^{\textrm{odd}}(n)$. The Alon-Tarsi conjecture states that $L^{\textrm{even}}(n)\neq L^{\textrm{odd}}(n)$ when $n$ is even (when $n$ is odd the two are equal for very simple reasons). In this short note we prove that $|L^{\textrm{even}}(n) - L^{\textrm{odd}}(n)|\leq L(n)^{\frac{1}{2} + o(1)},$ thus establishing the conjecture that the number of even and odd Latin squares, while conjecturally not equal in even dimensions, are equal to leading order asymptotically. Two proofs are given: both proceed by applying a differential operator to an exponential integral over $\mathrm{SU}(n)$. The method is inspired by a recent result of Kumar-Landsberg.