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.
Best Answer
If $\ X_i\ $ are independent standard normal variates then it's automatically true that $$ P(X_i=X_j)=\cases{1&if $\ i=j$\\ 0&if $\ i\ne j\ $,} $$ so your assumption that $\ P(X_i=X_j)=0\ $ for $\ i\ne j\ $ is redundant.
If $$ Q_r=T_{2r}=\sum_{i=1}^r(X_{2i-1}-X_{2i})^2\ , $$ then $\ Q_r\ $ is the sum of the squares of $\ r\ $ independent random variables $\ Y_i=X_{2i-1}-X_{2i}\sim N(0,2)\ $. Therefore $\ \frac{Q_r}{2}\ $ is the sum of the squares of $\ r\ $ independent standard normal variates, $\ \frac{Y_i}{\sqrt{2}}\ $. Thus, while $\ Q_r\ $ isn't, strictly speaking, $\ \chi^2\ $ distributed, $\ \frac{Q_r}{2}\ $ follows a chi-square distribution with $\ r\ $ degrees of freedom.