The numerator and denominator can also be written as $$E\left[\exp\left(\frac1n\sum\ln|X_i-X_j|\right)\right]$$ where the numerator has $n(n-1)/2$ summands and the denominator has $(n-1)(n-2)/2$ summands.
Let $\mu$ and $\sigma$ be the mean and standard deviation of $\ln|X_i-X_j|$. Since $X_i-X_j$ is a normal distribution with mean $0$ and standard deviation $\sqrt{2}$, $$\mu=E\left[\ln\Big|N(0,\sqrt{2})\Big|\right]=\frac{-\gamma}{2}$$ where $\gamma$ is the Euler-Mascheroni constant.
If the summands were independent, then the fraction in the question would be approximated by a ratio of means of lognormal distributions:
$$ \frac {E\left[LN\left(\frac{n(n-1)}{2n}\mu,\sqrt{\frac{n(n-1)}{2n^2}}\sigma\right)\right]} {E\left[LN\left(\frac{(n-1)(n-2)}{2n}\mu,\sqrt{\frac{(n-1)(n-2)}{2n^2}}\sigma\right)\right]} = \frac {\exp\left(\frac{n(n-1)}{2n}\mu+\frac{n(n-1)}{2n^2}\frac{\sigma^2}{2}\right)} {\exp\left(\frac{(n-1)(n-2)}{2n}\mu+\frac{(n-1)(n-2)}{2n^2}\frac{\sigma^2}{2}\right)} $$
In the limit this is $$ \exp\left(\frac{2(n-1)}{2n}\mu+\frac{2(n-1)}{2n^2}\frac{\sigma^2}{2}\right) \rightarrow \exp(\mu)=\exp(-\gamma/2) \sim 0.749.$$
Empirically the number seems to be between that and $1$.
With some numerical integration we could compute $\sigma$ and the correlation $\rho$ between $\ln|X_i-X_j|$ and $\ln|X_i-X_k|$; with some combinatorics we could compute how often those $\rho$'s show up in the standard deviations of the two lognormals; combining those results we could get a more accurate value for the overall expectation.