This is by no means a full answer, but points out some structure.
I assume that $\Omega$ is convex and regular enough.
Let us denote
$$
F(\Omega)
=
\int_{x\in\Omega}\int_{y\in\Omega}|x-y|\,dy\,dx.
$$
In the convex setting, you are minimizing $F$.
Let $\ell(x,v)$ denote the distance from $x\in\Omega$ to $\partial\Omega$ in the direction $v\in S^1$.
We use the convention that if $x\in\partial\Omega$ and $v$ points inward, then $\ell(x,v)>0$.
More analytically, we can define $\ell(x,v)=\sup\{t\geq0;x+tv\in\Omega\}$.
The inner integral defining $F(\Omega)$ can be written in polar coordinates.
The radial integral is easy to compute, and we find
$$
F(\Omega)
=
\frac13\int_{x\in\Omega}\int_{v\in S^1}\ell(x,v)^3\,dv\,dx.
$$
This is an integral over the ring bundle $S\Omega=\Omega\times S^1$ (also called the sphere bundle — $S^1$ is just the 1D sphere).
One can change integration from the sphere bundle an integral over all lines and the space of lines using the so-called Santaló formula (see e.g. proposition 8.2 in these lecture notes for a proof in $\mathbb R^2$).
This leads to
$$
F(\Omega)
=
\frac1{3}
\int_{x\in\partial\Omega}
\int_{v\in S^1}
|\langle v,\nu_x\rangle|
\int_0^{\ell(x,v)}\ell(x+tv,v)^3
\,dt\,dv\,dx.
$$
Here $\nu_x$ is the unit normal at $x$.
The innermost integral is again an explicit 1D integral, and we get
$$
F(\Omega)
=
\frac1{12}
\int_{x\in\partial\Omega}
\int_{v\in S^1_{x,in}}
|\langle v,\nu_x\rangle|
\ell(x,v)^4
\,dv\,dx.
$$
On the other hand, the area of $\Omega$ is
$$
|\Omega|
=
\frac1{2\pi}
\int_{x\in\partial\Omega}
\int_{v\in S^1_{x,in}}
|\langle v,\nu_x\rangle|
\ell(x,v)
\,dv\,dx.
$$
See exercise 96 in the linked notes.
Here $S^1_{x,in}$ is the set of $v\in S^1$ which point towards the interior of $\Omega$ from $x\in\partial\Omega$.
If $\partial_{in}S\Omega$ denotes the inward pointing boundary of the sphere bundle (the whole boundary is $\partial S\Omega=\partial\Omega\times S^1$) and we denote by $\sigma$ the measure corresponding to $|\langle v,\nu_x\rangle|\,dv\,dx$, we have
$$
F(\Omega)
=
\frac1{12}
\int_{\partial_{in}S\Omega}\ell(x,v)^4\,d\sigma(x,v)
$$
and
$$
|\Omega|
=
\frac1{2\pi}
\int_{\partial_{in}S\Omega}\ell(x,v)\,d\sigma(x,v).
$$
This puts $F(\Omega)$ and $|\Omega|$ in a very similar neat form.
The suspected extremal case is a disc of some radius $R>0$.
In this case $\ell(x,v)=2R|\langle v,\nu_x\rangle|$ when $v$ points inward.
Therefore it might be convenient to use the measure $\tilde\sigma$ corresponding to $dv\,dx$ instead.
Let us denote $u(x,v)=|\langle v,\nu_x\rangle|$ for brevity.
Let me denote the length of the perimeter by $P=|\partial\Omega|$ and the diameter of $\Omega$ by $D$.
Denoting $L^p=L^p(\partial_{in}S\Omega,\tilde\sigma)$, we have (for any $\alpha\geq0$ and $p\geq1$)
$$
\begin{split}
F(\Omega)&=\frac1{12}\|\ell^4u\|_{L^1},\\
|\Omega|&=\frac1{2\pi}\|\ell u\|_{L^1},\\
P&=\frac1\pi\|1\|_{L^p},\\
D&=\|\ell u^\alpha\|_{L^\infty}.
\end{split}
$$
These with the isoperimetric inequality $4\pi A\leq P^2$ and the isodiametric inequality $4A\leq\pi D^2$ and Hölder's inequality give something to play with.
To give anything sharp, one has to apply Hölder's inequality in a way that gives equality when $\ell$ is a constant multiple of $u$.
