Disclaimer - this treatment is not mathematical rigorous and I evaluate the
integral in a physicist manner. The end result is a formula for small $n$, one can derive the result with help of a CAS.
$\delta(x)$ isn't an ordinary function but a distribution. Since I don't know distribution, I will treat it as the limit of some sort of Gaussian peaks:
$$\delta(x) = \lim_{\epsilon\to 0} \delta_\epsilon(x)\quad\text{ where }\quad \delta_\epsilon(x) =\frac{1}{\sqrt{2\pi \epsilon^2} } e^{-\frac{x^2}{2\epsilon^2}}$$
I will also adopt physicists' way of writing integral.
Let $\Omega_n$ be your integral. Instead of $\Omega_n$, consider following integral
$$F_n(\lambda) = \int\prod_k d^3 x_k\;e^{-\lambda \sum_k|x_k|^2}\prod_{i<j}|x_i-x_j|^2 $$
Using following integral representation
$$e^{-\lambda \sum_k|x_k|^2} = \lim_{\epsilon\to 0}
\int_0^\infty dt e^{-\lambda t} \delta_\epsilon\left(\sum_k|x_k|^2-t\right)$$
We have
$$\begin{align}
F_n(\lambda) &= \lim_{\epsilon\to 0}\int_0^\infty dt e^{-\lambda t} \int
\prod_k d^3x_k \delta_\epsilon\left(\sum_k |x_k|^2 - t\right) \prod_{i<j}|x_i-x_j|^2\\
&= \lim_{\epsilon\to 0}\int_0^\infty dt e^{-\lambda t}
\int
\prod_k d^3x_k \frac1t\delta_\frac{\epsilon}{t}\left(\sum_k \frac{|x_k|^2}{t} - 1\right) \prod_{i<j}|x_i-x_j|^2
\end{align}
$$
Change variable to $y_k = \frac{x_k}{\sqrt{t}}$, we find
$$F_n(\lambda) = \lim_{\epsilon\to 0}\int_0^\infty dt\, t^{\frac{n(n+2)}{2} - 1} e^{-\lambda t} \int \prod_k d^3y_k \delta_\epsilon\left(\sum_k |y_k|^2 - 1\right)\prod_{i<j}|y_i - y_j|^2$$
and hence
$$\Omega_n = \frac{\lambda^{\frac{n(n+2)}{2}}}{\Gamma\left(\frac{n(n+2)}{2}\right)}F_n(\lambda)\tag{*1}
$$
Let $\beta_{ij} = \beta_{ji}$ for $1 \le i < j \le n$ be $\frac{n(n-1)}{2}$ parameters. Let $\alpha_k = \sum_{j\ne k} \beta_{kj}$ for $k = 1,\ldots, n$.
Consider following Gaussian integrals
$$G_n(\lambda,\beta_{ij}) \stackrel{def}{=} \int \prod_k d^3 x_k e^{-\lambda \sum_k |x|^2 + \sum_{i<j}\beta_{ij}|x_i - x_j|^2}
= \pi^{\frac{3n}{2}} \Lambda_n(\lambda-\alpha,\beta)^{-3/2}
$$
where $\Lambda_n(\lambda-\alpha,\beta)$ is the determinant
$$\left|
\begin{matrix}
\lambda - \alpha_1 & \beta_{12} & \beta_{13} & \cdots\\
\beta_{12} & \lambda - \alpha_2 & \beta_{23} & \cdots\\
\vdots & \vdots & \ddots & \vdots\\
\beta_{1\,n-1} & \beta_{2\,n-1} & \cdots & \lambda - \alpha_n
\end{matrix}
\right|$$
It is easy to see
$\displaystyle\;F_n(\lambda) = \left. \left(\prod_{i<j} \frac{\partial}{\partial \beta_{ij}}\right)
G_n(\lambda,\beta_{ij})\right|_{\beta = 0}\;$
Plug this into $(*1)$ and set $\lambda$ to $1$, we obtain a formula for $\Omega_n$.
$$\Omega_n = \frac{\pi^{\frac{3n}{2}}}{\Gamma\left(\frac{n(n+2)}{2}\right)}
\left.\left(\prod_{i<j} \frac{\partial}{\partial \beta_{ij}}\right) \Lambda_n(1 - \alpha, \beta)^{-3/2}\right|_{\beta=0}$$
As a double check, let us evaluate this for $n = 2$ and $3$, we find
$$\begin{align}
\Omega_2 &
= \frac{\pi^3}{\Gamma(4)}\left.\frac{\partial}{\partial \beta}
\left|\begin{matrix}
1 - \beta & \beta \\
\beta & 1 - \beta
\end{matrix}\right|^{-3/2}\right|_{\beta=0}\\
& = \frac{\pi^3}{3!}\times 3 = \frac{\pi^3}{2}\\
\Omega_3
&= \frac{\pi^{\frac92}}{\Gamma(\frac{15}{2})}\left.\frac{\partial^3}{\partial u\partial v\partial w}
\left|\begin{matrix}
1 - u - v & u & v \\
u & 1 - u - w & w \\
v & w & 1 - v - w
\end{matrix}\right|^{-3/2}\right|_{u=v=w=0}\\
&= \frac{\pi^{\frac92}}{\Gamma(\frac{15}{2})}\times \frac{75}{2} = \frac{320\pi^4}{9009}
\end{align}
$$
which reproduce the values of $\Omega_2$ and $\Omega_3$ I obtained by evaluating the original integrals by hand.
Let me repeat once again: this is not mathematical rigorous. Even from a physicist standpoint, if one want to use this, one should at least compute $\Omega$ for other small $n$ by hand or Monte Carlo and see whether this formula produces correct number.
For larger $n$, one probably need to derive a diagrammatic expansion of above determinant to make evaluation of above formula practical.
