Mathematica can do these integrals explicitly, for small $n$. One gets
$$
\frac1{n!}\int_{\Omega_1^{(n)}}\prod_{1\leq i<j\leq n}(x_i-x_j)^2
\prod_{i=1}^ne^{-x_i^2}\,dx = c_n\,\pi^{\frac{n-1}2}
e^{-\lambda^2}\lambda^{2n-3}\left(1+O\left(\frac1{\lambda^2}\right)\right) \tag{1}
$$
with $c_2=\frac14$, $c_3=\frac1{12}$, $c_4=\frac1{30}$, $c_5=\frac3{160}$, and $c_6=\frac{3}{128}$.

Here is a **rough upper estimate** which reproduces this result, but with too large
a constant $c_n'$ (replacing $c_n\,\pi^{\frac{n-1}2}$): Write $x'=(x_2,\dots,x_n)$ with $r=|x'|$ and use
$$
\prod_{1\leq i<j\leq n}(x_i-x_j)^2\leq 2^{(n-2)(n-1)}(|x_1|+r)^{2n-2}r^{(n-2)(n-1)}.
\tag{2}
$$
Then, for any $\lambda>0$,
$$
\begin{aligned}
\int_{\Omega_1^{(n)}}\prod_{1\leq i<j\leq n}(x_i-x_j)^2
\prod_{i=1}^ne^{-x_i^2}\,dx &\leq
\int_\lambda^\infty\int_{{\mathbb R}^{n-1}}\prod_{1\leq i<j\leq
n}(x_i-x_j)^2 \prod_{i=1}^ne^{-x_i^2}\,dx'dx_1 \\ &\leq
\frac{2^{(n-2)(n-1)}n\,\pi^{\frac{n}2}}{\Gamma\left(\frac{n+2}2\right)}\,
\int_\lambda^\infty\int_0^\infty(x_1+r)^{2n-2}r^{n(n-2)}
e^{-x_1^2-r^2}\,drdx_1.
\end{aligned}
$$
Now,
$$
\begin{aligned}
\int_0^\infty (x+r)^{2n-2}r^{n(n-2)} e^{-x^2-r^2}\,dr
&\leq 2^{2n-3}e^{-x^2}\left(x^{2n-2}
\int_0^\infty
r^{n(n-2)} e^{-r^2}\,dr + \int_0^\infty r^{n^2-2} e^{-r^2}\,dr\right) \\
&\leq 2^{2n-4}e^{-x^2} \left(\Gamma\left(\frac{n^2-2n+1}2\right)x^{2n-2}
+\Gamma\left(\frac{n^2-1}2\right)\right)
\end{aligned}
$$
and, therefore,
$$
\begin{aligned}
& \int_\lambda^\infty \int_0^\infty (x+r)^{2n-2}r^{n(n-2)}
e^{-x^2-r^2}\,drdx \\
& \qquad \leq 2^{2n-4}\int_\lambda^\infty e^{-x^2}
\left(\Gamma\left(\frac{n^2-2n+1}2\right)x^{2n-2}
+\Gamma\left(\frac{n^2-1}2\right)\right) dx \\ & \qquad =
2^{2n-5}\left(\Gamma\left(\frac{2n-1}2,\lambda^2\right)
\Gamma\left(\frac{n^2-2n+1}2\right) +
\Gamma\left(\frac12,\lambda^2\right)
\Gamma\left(\frac{n^2-1}2\right)\right).
\end{aligned}
$$
Next one needs to bound the incomplete $\Gamma$ function $\Gamma(\alpha,\mu)$ as $\alpha\to\infty$ uniformly in the parameter $\mu$. Such estimates can be found here.

**Note.** 1. Knowing the asymptotics
$$
\int_\lambda^\infty \int_0^\infty (x+r)^{2n-2}r^{n(n-2)} e^{-x^2-r^2}\,dr dx =
\frac14\,\Gamma\left(\frac{(n-1)^2}2\right)
e^{-\lambda^2}\lambda^{2n-3} \left(1+O\left(\frac1\lambda\right)\right)
$$
as $\lambda\to\infty$ is less effective, as the dependence of the remainder on $n$ is not exhibited.

Without having done these computations, I would say that one has to be
**more sensitive in estimate (2)**. For instance, expanding $\prod_{1\leq i<
j\leq n}(x_i-x_j)^2$, all the terms $\prod_{i=1}^n x_i^{k_i}$ (notice
that $\sum_{i=1}^nk_i=n(n-1)$) with (at least) one $k_i$ being odd contribute nothing to the integral over $(\lambda,\infty)\times{\mathbb R}^{n-1}$. **Once the combinatorics is done, the method as outlined should work.**

With more effort, I belief that one can work out (1) for all $n$ (including a formula for the $c_n$). Still, this does not settle the question on the dependence of the remainder on $n$.

Flipping over one $x_i<\lambda$ to $x_i>\lambda$ introduces another factor $c\,e^{-\lambda^2}\lambda^N$ in the asymptotics as $\lambda\to\infty$, for some constant $c>0$ and integer $N$ (with varying $c$, $N$). This probably can be worked out, too, but it will get messy rather quickly if one wants to know these $c$, $N$.