Expected triangle area of normal distributed vertices with colinear expectations Situation
Given are 3 independent multinormal distributions $X_i=\mathcal{N}(\vec\mu_i,\Sigma)_{i=1,2,3}$ in $\mathbb{R^3}$.
For simplification the expectations are colinear:
$\vec\mu_1=\begin{pmatrix}-a\\0\\0\end{pmatrix}, \vec\mu_2=\begin{pmatrix}0\\0\\0\end{pmatrix}, \vec\mu_3=\begin{pmatrix}b\\0\\0\end{pmatrix}$ with $a\ge0, b\ge0$.
The covariance matrix is:
$\Sigma=\begin{pmatrix}\sigma^2&0&0\\0&\sigma^2&0\\0&0&\sigma^2\end{pmatrix}$.
Question
What is the expected absolute area $\mathbb{E}(A)$ of triangle $x_1,x_2,x_3$ with $x_i\sim~X_i$?
Sketch

No right solution was given in two bounties
The answerer of the 1st bounty solved the easier problem $\mathbb{E}(A^2)$ but not $\mathbb{E}(A)$. No answer was given in the 2nd bounty.

What is known?
Simplified cases: solutions and approximations

*

*$\,\,\,\,\,\,\text{max}(a,b)=0 \rightarrow \mathbb{E}(A)=\sqrt{3}\sigma^2$
(proof)


*$\,\,\,\,\,\,a \gg \sigma \land b=0 \rightarrow \mathbb{E}(A)=\frac{\sigma}{2}\sqrt{\pi}a$ (proof below)


*$\,\,\,\,\,\,\text{max}(a,b) \ll \sigma \rightarrow \mathbb{E}(A) \approx \sqrt{3}\sigma^2$ (presumed by simulations)


*$\,\,\,\,\,\,\text{min}(a,b) \gg \sigma \rightarrow \mathbb{E}(A)\approx \frac{\sigma}{2}\sqrt{\pi(a^2+b^2+ab)}$ (presumed by simulations)
Proof for case 2:
As $a\gg \sigma$ the triangle can be assumed as a right triangle and the expected area is $\mathbb{E}(A)=\frac{1}{2}ac$ with length $c=pq$. The length $p$ is derived from the expectation of a shifted central chi-distribution with 3 degrees of freedom: $\frac{p}{\sqrt{2}\sigma}= \mathbb{E}(\chi_3)=\sqrt{\frac{8}{\pi}}$. The length must be corrected by $q=\frac{\pi}{4}$, the perpendicular component of $p$ to the line $\overline{\mu_1 \mu_2}$ (see here).

 A: I'll try to give a partial answer since the bounty is expiring soon.
Let me call the 3 components of the $X_i$ to be $x_i$, $y_i$, and $z_i$
The norm of the cross product of two vectors is the area of the parallelogram swept by them.  So half of that is the triangle you're looking for.
$\text{area}=\frac{1}{2} \left\| \left(\left\{x_2,y_2,z_2\right\}-\left\{x_1,y_1,z_1\right\}\right)\times \left(\left\{x_3,y_3,z_3\right\}-\left\{x_1,y_1,z_1\right\}\right)\right\|$
Simplifying:
$\text{area}=\frac{1}{2} \sqrt{\left(x_3 \left(y_2-y_1\right)+x_2 \left(y_1-y_3\right)+x_1 \left(y_3-y_2\right)\right){}^2+\left(x_3 \left(z_2-z_1\right)+x_2 \left(z_1-z_3\right)+x_1 \left(z_3-z_2\right)\right){}^2+\left(y_3 \left(z_2-z_1\right)+y_2 \left(z_1-z_3\right)+y_1 \left(z_3-z_2\right)\right){}^2}$
The square of the area is much nicer to work with:
$\text{areaSq}=\frac{1}{4} \left(\left(x_3 \left(y_2-y_1\right)+x_2 \left(y_1-y_3\right)+x_1 \left(y_3-y_2\right)\right){}^2+\left(x_3 \left(z_2-z_1\right)+x_2 \left(z_1-z_3\right)+x_1 \left(z_3-z_2\right)\right){}^2+\left(y_3 \left(z_2-z_1\right)+y_2 \left(z_1-z_3\right)+y_1 \left(z_3-z_2\right)\right){}^2\right)$
So now that we know the area in terms of the random variables, the expectation becomes a definite integral over the domains of the three random variables.  Mathematica can easily integrate this.

which works out to:
$\frac{1}{2} \left(2 a^2 \sigma ^2+2 a b \sigma ^2+2 b^2 \sigma ^2+9 \sigma ^4\right)$
and then we can apply Jensen's inequality which supports your assertions about asymptotic cases.
Unfortunately, trying to find the expectation of the absolute value of area rather than the square of area results in a super nasty integral with an analytic solution that evidently exceeds my patience and my computer's 32 GB of memory.
I'm not surprised that you see a $\sqrt{\pi }$ term in your empirically observed asymptotic limits since in the very simple case of looking at the expected value of the absolute value of a normally distributed random variable:
$\text{Expectation}[\left| x\right| ,x\text{~NormalDistribution}[\mu ,\sigma ]]$
evaluates to the rather ugly: 
$\mu  \text{ Erf}\left(\frac{\mu }{\sqrt{2} \sigma }\right)+\sqrt{\frac{2}{\pi }} \sigma  e^{-\frac{\mu ^2}{2 \sigma ^2}}$
Hopefully this is enough to inspire you to get the answer you need!
