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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.

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.

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 $\{\chi_{1},\chi_{2}, \ldots \chi_{t}$ are 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.

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.

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 $\{\chi_{1},\chi_{2}, \ldots \chi_{t}$ are 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 $i = \sqrt{-1}.$

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 $\{\chi_{1},\chi_{2}, \ldots \chi_{t}$ are 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.