Let $\eta=e^{\frac{2\pi i}n}$, an $n$-th root of unity. For pedagogical reasons and inspiration, I ask to see different proofs (be it elementary, sophisticated, theoretical, etc) for the following product evaluation.
If $T(n)=\frac{(3n-2)(n-1)}2$ and $i=\sqrt{-1}$ then $$\prod_{j<k}^{0,n-1}(\eta^k-\eta^j)=n^{\frac{n}2}i^{T(n)}.$$
We first find the norm; we then determine the argument.
Call the product you wrote $A_n$. Then $A_n^2 = \prod_{j<k}^{0,n-1} (\eta^k - \eta^j)^2 = Disc(x^n - 1) = (-1)^{\frac{n (n -1)}{2}}Res(x^n - 1, n x^{n - 1})$
$= (-1)^{\frac{n(n-1)}{2}} n^n \prod_{0 \leq i < n, 0 \leq j < n-1} (\eta^i - 0)$
All terms in the expression except $n^n$ have norm $1$, so we have that $|A_n| = n^{\frac{n}{2}}$. We therefore only need to figure out the argument of $A_n$.
Let $\eta' = e^\frac{2 \pi i}{2n}$ be the square root of $\eta$. We can rewrite $A_n = \prod_{0\leq j<k<n} \eta'^{k + j} (\eta'^{k - j} - \eta'^{j - k})$. Note that the second term is a difference of (unequal) conjugates where the minuend has positive imaginary part (and the subtrahend therefore negative imaginary part), and therefore will always have argument $\frac{\pi}{2}$. So let us concentrate on the argument of the first term, $\prod_{0 \leq j < k < n} \eta'^{k +j}$. We can do this by finding $\sum_{0 \leq j < k < n} j + k$.
$\sum_{0 \leq j < k < n} j + k = \left(\sum_{0 \leq j < k < n} j\right) + \left(\sum_{0 \leq j < k < n} k\right)$
$= \left(\sum_{0 \leq j <n} (n - j - 1)j\right) + \left(\sum_{0\leq k<n} k*k\right)$
$= \sum_{0 \leq j < n} (n - j - 1)j + j*j = \sum_{0 \leq j < n} (n - 1)j$
$= (n - 1) \frac{n (n - 1)}{2}$
We therefore end up with an argument of $\frac{n(n - 1)}{2} \frac{\pi}{2} + \frac{n (n - 1)^2}{2} \frac{2 \pi}{2n} = \frac{(3n^2 - 5n + 2)\pi}{4}$. We finally have that:
The norm of $A_n$ is $n^\frac{n}{2}$, and the argument is $\frac{(3n^2 - 5n + 2)\pi}{4}$. Correspondingly, we have that $A_n = n^{\frac{n}{2}} i^{T(n)}$, as desired.
Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.
Your determinant is essentially the Van der Monde determinant ${\rm det}(A),$ where $A$ is the $n \times n$ matrix $[\eta^{(j-1)(k-1)}].$
Note that $A$ is the character table of the cyclic group of order $n$ so that $A{\bar A}^{T}= nI_{n}$, using the orthogonality relations for group characters, and $|{\rm det A}| = n^{\frac{n}{2}}.$ One can continue in this vein, but I will sketch a more general calculation of the determinant of the character table of a general finite group, which simplifies considerably in the case of cyclic groups.
Let $G$ be a finite group with $t$ conjugacy classes, say with representatives $g_{1},g_{2}, \ldots g_{t}.$ Let us label these classes so that $1_{G} = g_{1}, g_{2},\ldots ,g_{s}$ are exactly those class representatives which are conjugate to their inverses in $G$ and so that $g_{s+2j} = g_{s+2j-1}^{-1}$ for $1 \leq j \leq \frac{t-s}{2}.$
Let $\chi_{1},\chi_{2}, \ldots \chi_{t}$ be the complex irreducible characters of $G.$
Let $B$ be the character table of $G$, which is the $t \times t$ matrix $[\chi_{j}(g_{k})].$
By the orthogonality relations for group characters, we see that ${\bar B}^{T}B$ is the diagonal matrix whose $j$-th diagonal entry is $|C_{G}(g_{j})|.$
Let $\pi \in {\rm S}_{t}$ be the permutation fixing $1,2,\ldots,s$ and interchanges $s+2j-1$ and $s+2j$ for $1 \leq j \leq \frac{t-s}{2}.$ Let $P$ be the associated permutation matrix.$
Note that $BP = {\bar B}$ since $BP$ has the same first $s$ columns as $B$ and has the columns corresponding to $g_{s+2j-1}$ and $g_{s+2j} = g_{s+2j-1}^{-1}$ interchanged for $1 \leq \frac{t-s}{2}$ (note that the first $s$ columns of $B$ are real as $g_{j}$ is conjugate to $g_{j}^{-1}$ for $1 \leq j \leq s.$
Hence ${\rm det}B^{2} = (-1)^{\frac{t-s}{2}} \prod_{j=1}^{t} |C_{G}(g_{j})|$ and ${\rm det}B = (i)^{\frac{t-s}{2}} \sqrt{\prod_{j=1}^{t} |C_{G}(g_{j})|}$ where $i = \sqrt{-1}.$
Edit following Mark Wildon's comment: In fact, although it looks as though there is a free choice of $\sqrt{-1}$ in the above, in the case that $G$ is cyclic of order $n$, if we define $\eta$ as in the question as $\eta = \exp{\frac{2 \pi i}{n}},$ then we have to use the other square root of $-1$ in the above expression for ${\rm det}(A),$ so ${\rm det}A = (-i)^{\frac{n-s}{2}} n^{\frac{n}{2}}$ where $s =1 $ if $n$ is odd and $s = 2$ if $n$ is even.
Even later edit: Here are some remarks about the general character table determinant which can be seen directly without making a choice for $\sqrt{-1}.$ Note that since ${\bar B} = BP$ and $\overline{{\rm det}B} = (-1)^{\frac{t-s}{2}}{\rm det}B$, we have that ${\rm det}(B) \in \mathbb{R}$ if and only if $t \equiv s$ (mod $4$). When $t \not \equiv s$ (mod $4$), we see that ${\rm det}B$ is pure imaginary. At present, I don't see a quick way (in the general case) to determine which choice of $\sqrt{-1}$ to use in the earlier formula once a choice of $i$ is fixed for the character table entries.
Here is a proof using the logarithmic function $\mathrm{Li}_1(z)=-\log(1-z)$:
Let $P= \displaystyle\prod_{\substack{j,k=0 \\ j<k}}^{n-1} (\eta^k-\eta^j)$. Take the logarithm: \begin{align*} \log P & = \sum_{j<k} \log \eta^k - \sum_{j<k} \mathrm{Li}_1(\eta^{j-k}) \\ & = \sum_{k=0}^{n-1} k \cdot \frac{2\pi i k}{n} - \sum_{a=1}^{n-1} (n-a) \mathrm{Li}_1(\eta^{-a}) \\ & = \frac{2\pi i}{n} \sum_{k=0}^{n-1} k^2 - \sum_{a=1}^{n-1} a \mathrm{Li}_1(\eta^a). \end{align*} Call $S$ the second sum. We have \begin{align*} S & = \sum_{a=1}^{n-1} a \mathrm{Li}_1(\eta^a) = \frac12 \sum_{a=1}^{n-1} \bigl(a \mathrm{Li}_1(\eta^a)+(n-a)\mathrm{Li}_1(\eta^{-a})\bigr)\\ & = \frac{n}{2} \sum_{a=1}^{n-1} \mathrm{Li}_1(\eta^a) + \sum_{a=1}^{n-1} a \cdot \bigl(\mathrm{Li}_1(\eta^a)-\mathrm{Li}_1(\eta^{-a})\bigr) \end{align*} The first sum is easy to compute and is equal to $-\log n$. Regarding the second sum, the classical Fourier expansion for the Bernoulli polynomial $B_1(x)=x-\frac12$ on $(0,1)$ gives \begin{equation*} \mathrm{Li}_1(\eta^a)-\mathrm{Li}_1(\eta^{-a}) = 2\pi i \bigl(\frac12 - \frac{a}{n}\bigr). \end{equation*} From there, it is not difficult to finish the computation \begin{align*} \log P = \frac{n}{2} \log n - \frac{\pi i}{4} n(n-1) + \frac{3\pi i}{n} \sum_{k=1}^{n-1} k^2 = \frac{n}{2} \log n + \frac{\pi i}{4} (n-1)(3n-2) \end{align*} which gives the desired value for $P$.
