Two equivalent formulas for the volume of a random tetrahedron are given. Further on you can find an interesting conjecture for the expected volume that shall be proved.
Tetrahedron volume
Given are 12 independent standard normal distributed variables
$$x_i=\mathcal{N}(0,1)_{i=1,…,12}$$
that define the 4 coordinates
$$\vec{a}=(x_1,x_2,x_3),\;\; \vec{b}=(x_4,x_5,x_6),\;\; \vec{c}=(x_7,x_8,x_9),\;\; \vec{d}=(x_{10},x_{11},x_{12})$$
of a 3-simplex in $\mathbb{R}^3$. The first formula for the non-oriented simplex volume is
$$V=\frac{1}{6}\left| (\vec{a}-\vec{d})\cdot \left((\vec{b}-\vec{d}) \times (\vec{c}-\vec{d})\right) \right|\tag{1}$$
$$=\frac{1}{6}\left| x_2 x_6 x_7 + x_3 x_4 x_8+ x_1 x_5 x_9+ x_3 x_5 x_{10} + x_6 x_8 x_{10} + x_2 x_9 x_{10}+ x_1 x_6 x_{11}+ x_3 x_7 x_{11}+ x_4 x_9 x_{11}+ x_2 x_4 x_{12}+ x_5 x_7 x_{12}+ x_1 x_8 x_{12}-x_3 x_5 x_7- x_2 x_6 x_{10}- x_3 x_8 x_{10} – x_1 x_6 x_8 – x_2 x_4 x_{9}- x_5 x_9 x_{10}- x_3 x_4 x_{11}- x_6 x_7 x_{11}- x_1 x_9 x_{11}- x_1 x_5 x_{12}- x_2 x_7 x_{12}- x_4 x_8 x_{12}\right|.$$
If the coordinate system is shifted
$$\vec{p}=\vec{a}-\vec{d},\;\;\vec{q}=\vec{b}-\vec{d},\;\;\vec{r}=\vec{c}-\vec{d}$$
the new coordinates are
$$\vec{p}=(y_1,y_2,y_3),\;\; \vec{q}=(y_4,y_5,y_6),\;\;\vec{r}=(y_7,y_8,y_9)$$
with new random variables
$$y_i=\mathcal{N}(0,\sqrt{2})_{i=1,…,9}.$$
The shift reduces the number of random variables from 12 to 9 and increases the standard deviation from $1$ to $\sqrt{2}$ (this corresponds to a double variance $=\sqrt{2}^2)$. However the variables are not independent anymore.
Their correlation $\rho=0.5$ is given by their covariance normalized by the standard deviation
$$\rho=\frac{\mathbb{Cov}[y_i,y_j]}{\sqrt{\mathbb{Var}[y_i]}\sqrt{\mathbb{Var}[y_j]}}=
\frac{\mathbb{Cov}[x_m-x_k,x_n-x_k]}{\sqrt{\mathbb{Var}[x_m-x_k]}\sqrt{\mathbb{Var}[x_n-x_k]}}
=\frac{\mathbb{E}[x_k^2]}{\sqrt{\mathbb{Var}[x_m-x_k]}\sqrt{\mathbb{Var}[x_n-x_k]}}=\frac{\mathbb{E}[x_k]^2+\mathbb{Var}[x_k]}{\sqrt{\mathbb{Var}[x_m-x_k]}\sqrt{\mathbb{Var}[x_n-x_k]}}
=\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}\;\;\;\text{for}\;i\ne j \land n\ne m \ne k.$$
The second formula for the non-oriented volume as function of the dependent variables is
$$V=\frac{1}{6}\left|\vec{p}\cdot (\vec{q} \times \vec{r}\right)|\tag{2}$$
$$=\frac{1}{6}\left| y_2y_6y_7+y_3y_4y_8+y_1y_5y_9-y_1y_6y_8-y_2y_4y_9-y_3y_5y_7\right|.$$
Equation (2) has only a quarter of summands of eq.(1) however the variables correlate with $\rho=0.5$.
Question
What is the analytical expression for the expected volume $\mathbb{E}[V]$?
What is known?
Conjecture
It is conjectured that $\mathbb{E}[V]=\frac{2}{3}\sqrt{\frac{2}{\pi}}$ or $\mathbb{E}[V]=\frac{21}{4\pi^2}$. Assuming the first conjecture is true please note the relation to a standard half-normal distribution in $\mathbb{R^1}$ that has expectation $\sqrt{\frac{2}{\pi}}$.
Moments
All even moments are precisely known and the odd moments are approximately known. The first moments are
\begin{array}{|l|l|}\hline
\text{odd moments} & \text{even moments} \\
\text{(simulation)} & \text{(analytic)} \\ \hline
m_1\approx 0.532 & m_2=\frac{2}{3}\\ \hline
m_3\approx\sqrt{2} &m_4=\frac{40}{9} \\ \hline
m_5\approx18.9 &m_6=\frac{2800}{27} \\ \hline
\end{array}
(more moments on demand).
Solution strategies
One could try to integrate over a subvolume where the sign of the volume is constant. Due to symmetry every subvolume should have equal size. The challenge is therefore to find the right suitable integration borders.
A related question about the expected area of a triangle with standard normal distributed coordinates in $\mathbb{R}^3$ was proven to be $\sqrt{3}$. If these methods would be applied to the tetrahedron case then according to the answerer "ultimately it comes down to the product of independent chi-distributed variables and a variable for the spherical angle they determine: finding the expectation of the latter is the crux of the question."
Other equations for the volume
There are other methods to calculate the volume however they include at least 1 square root, an unwanted property for such problems.
Expected oriented volume
The expression for the volume is a sum of triple products of random variables. As the expectations of the independent $x_i$ in eq.(1) are $\mathbb{E}[x_i]=0$ it holds
$$\mathbb{E}[x_i x_j x_k\pm x_l x_m x_n]=0\cdot 0 \cdot 0\pm 0\cdot 0 \cdot 0=0\;\;\;\text{for}\; 1\le i,j,k,l,m,n \le 12$$
The expected oriented volume is therefore $0$.
Best Answer
Let $X_0, X_1, \dots, X_n$ be i.i.d. standard normal vectors in $\mathbb{R}^n$ (so each $X_i \sim \mathcal{N}(0, I_n)$). Writing $Y_i = X_i - X_0$ for $i = 1, \dots, n$, we have that the $n$-volume of the $n$-simplex with vertices $X_0, X_1, \dots, X_n$ is equal to $$\frac{1}{n!} |\det(Y_1, \dots, Y_n)|$$ where we consider $Y_1, \dots, Y_n$ as column vectors.
Define $(W_1, \dots, W_n) = (Y_1, \dots, Y_n)^T$, i.e. $W_{i, j} = X_{j, i} - X_{0, i}$, so $W_1, \dots, W_n$ are independent, and $W_i \sim \mathcal{N}(0, \Sigma)$, where the covariance matrix $\Sigma$ has $2$'s on the diagonal and $1$'s off the diagonal. Note that $J_n$ (the matrix of ones) has eigenvalues $n, 0, \dots, 0$, hence since $\Sigma = I_n + J_n$, $\Sigma$ has eigenvalues $n+1, 1, \dots, 1$ and thus $\det \Sigma = n+1$. Now, defining $Z_i = \Sigma^{-1/2} W_i$ for $i = 1, \dots, n$, we have that $Z_1, \dots, Z_n$ are independent with each $Z_i \sim \mathcal{N}(0, I_n)$, and also that $$\det(Y_1, \dots, Y_n) = \det(W_1, \dots, W_n) = \det(\Sigma^{1/2}Z_1, \dots, \Sigma^{1/2}Z_n) = \det \Sigma^{1/2} \cdot \det(Z_1, \dots, Z_n).$$ It follows that the desired expected volume is $$\frac{\sqrt{n+1}}{n!} \mathbb{E}[|\det(Z_1, \dots, Z_n)|]$$ for independent $Z_1, \dots, Z_n \sim \mathcal{N}(0, I_n)$. To finish, we compute $\mathbb{E}[|\det(Z_1, \dots, Z_n)|]$.
Let $Z_1', \dots, Z_n'$ be the result of performing the Gram-Schmidt process to $Z_1, \dots, Z_n$ without normalizing, so for each $k$, we have $\mathrm{span}(Z_1', \dots, Z_k') = \mathrm{span}(Z_1, \dots, Z_k)$, and we inductively define $Z_k' = Z_k - P_kZ_k$ (with $Z_1' = Z_1$), where $P_k$ is the orthogonal projection onto $\mathrm{span}(Z_1', \dots, Z_{k-1}')$. Notably, these are all elementary column operations, so $\det(Z_1', \dots, Z_n') = \det(Z_1, \dots, Z_n)$, and $Z_1', \dots, Z_n'$ are orthogonal, so $|\det(Z_1', \dots, Z_n')| = \prod_{k=1}^n |Z_k'|$. Equivalently, we have $Z_k' = P_k' Z_k$, where $P_k'$ is the orthogonal projection onto the orthogonal complement of $\mathrm{span}(Z_1', \dots, Z_{k-1}')$, so $Z_k'$ can be seen as a standard normal vector on this $(n-k+1)$-dimensional space. This means that conditioning on $Z_1', \dots, Z_{k-1}'$, $|Z_k'|$ has the chi distribution with $n-k+1$ degrees of freedom, so in fact $|Z_k'|$ is independent of $Z_1', \dots, Z_{k-1}'$ with $$\mathbb{E}[|Z_k'|] = \sqrt{2} \frac{\Gamma((n-k+2)/2)}{\Gamma((n-k+1)/2)}.$$ It follows that all $|Z_k'|$ are independent, giving \begin{align*} \mathbb{E}[|\det(Z_1, \dots, Z_n)|] &= \prod_{k=1}^n \mathbb{E}[|Z_k'|]\\ &= \prod_{k=1}^n \sqrt{2} \frac{\Gamma((n-k+2)/2)}{\Gamma((n-k+1)/2)} \\ &= \prod_{k=1}^n \sqrt{2} \frac{\Gamma((k+1)/2)}{\Gamma(k/2)} \\ &= 2^{n/2} \frac{\Gamma((n+1)/2)}{\Gamma(1/2)} \end{align*} so the expected volume is $2^{n/2} \frac{\Gamma((n+1)/2) \sqrt{n+1} }{\Gamma(1/2) n!}$. At $n = 3$ (the given case), this is $\frac{2}{3} \sqrt{\frac{2}{\pi}}$.
Higher moments can be computed in the same way, using the corresponding higher moments of the chi distribution.