1) I tried to write down a proof of the above result. Let $M/L/K$ be Galois extensions, and the Galois groups $G=\mathrm{Gal}(M/K)$ and $H=\mathrm{Gal}(M/L)$. Then $L=M^H$ and $H$ is normal in $G$ and the quotient $G/H$ is the Galois group of the Galois extension $M/L$. Let $G/H=\{\sigma_1H,\dots,\sigma_mH\}$ for $\sigma_i\in G$. Let $a$ generate a $K$-normal basis for $M$, then the sum $s=\mathrm{Tr}_{M/L}(a)=\sum_{h_i\in H}h_i(a)$ is an element of $M^H=L$. To show that $s$ generates a normal $k$-basis of $L$, we take a $k$-linear dependance relation $0=\sum_i b_i\sigma_i(s)$ so $$0=\sum_i b_i\sigma_i(\sum_j h_j(a))=\sum_i\sum_j b_i(\sigma_ih_j)(a)=\sum_{g\in G}b_g g(a)$$ since the double sum collects all $G$, so $b_g=0$ since $a$ generates a normal basis. Hence $s$ generates a normal $K$-basis of $L$.
2) Application. Following you, $c=2\cos(2\pi/p)$ generates a normal basis. Let $K=\mathbb{Q}$ and consider $M=K(\zeta)$ with $\zeta=\exp(i2\pi/p)$. Then $\{\zeta,\zeta^2,\dots,\zeta^{p-1}\}$ is a normal $K$-basis of $M$: if $a_1\zeta+\cdots+a_{p-1}\zeta^{p-1}=0$ implies $a_1+a_2\zeta+\cdots+a_{p-1}\zeta^{p-2}=0$ so $a_i=0$ since $1,\zeta,\dots,\zeta^{p-2}$ is a power basis of $M$. Then consider $c=\zeta+1/\zeta \in \mathbb{R}$, so $L=K(c)$ is a real subfield of $M\subset \mathbb{C}$ so $M/L$ is Galois. The minimal of $\zeta$ over $L$ is $T^2-cT+1=(T-\zeta)(T-1/\zeta)$ so the Galois group is $H=G(M/L)=\{x\mapsto x,x\mapsto 1/x\}$. Since $G=\mathrm{Gal}(M/K)$ is a cyclic group, thus abelian, thus $H$ is a normal subgroup of $G$, and $L=M^H$, so the extension $L/K$ is Galois by the correspondance. Hence we can use the result above: $\mathrm{Tr}_{M/L}(\zeta)=\zeta+1/\zeta=c=2\cos(2\pi/p)$ generates a normal $K$-basis of $L$.
Write $p$ as $2q+1$, so $p\equiv 1\mod 4$ when $q$ is even, and $p\equiv 3\mod 4$ otherwise. The discriminant of the cyclotomic polynomial $\Phi_p$ is $\prod_{i<j}(\zeta^i-\zeta^j)^2=(-1)^{p(p-1)/2}p^{p-2}$ $>0$, so $\sqrt{p}$ generates a subfield $F=K(\sqrt{p})\subset L=K(c)$ when $p\equiv 1$ and we have a tower $L/F/K$ of Galois fields of degrees $(p-1)/4$ et $2$. Then by the same result, since $L/F/K$ is a Galois tower, $\mathrm{Tr}_{L/F}(c)=r+s\sqrt{p}\in F$ generates a $K$-normal basis $\{r\pm s\sqrt{p}\}$ for $F$.
3) I was initially interested in the determinant of the matrice $[g_ig_j(c)]$ (where $g_i\in G(L/K)$). This matrice checks if $c$ generates a normal basis. I wanted to understand why it is not an integer in case when $p\equiv 1\mod 4$. The final result seems to be that $\det[g_ig_j(c)]=-\sqrt{p}^{p-3}$
Once the automorphisms of $G(L/K)$ are determined (in the example for $p=13$ below), one computes the permutations of the orbit of $c$ under $G(L/K)$, then obtain the symmetric matrix $M(c)=[g_ig_j(c)]$. For the case $p=13$ for example, one obtains: $$M(c)=\begin{pmatrix}c_{1} & c_2 & c_3 & c_4 & c_5 & c_6\\ c_2 & c_4 & c_6 & c_5 & c_3 & c_1\\ c_3 & c_6 & c_4 & c_1 & c_2 & c_5\\ c_4 & c_5 & c_1 & c_3 & c_6 & c_2\\ c_5 & c_3 & c_2 & c_6 & c_1 & c_4\\ c_6 & c_1 & c_5 & c_2 & c_4 & c_3 \end{pmatrix}\begin{array}{c} () &\sigma^6 &(+)\\ (124536) &\sigma &(-)\\ (134)(265) &\sigma^4 &(+)\\ (143)(256) &\sigma^2 &(+)\\ (15)(23)(46) &\sigma^3 &(-)\\ (163542) &\sigma^5 &(-) \end{array}$$ and these permutations form a cyclic subgroup of $S_6$ of order $6$. We compute the matrix $D=M\cdot M^t$. It has the form: $$M\cdot M^t=\left(\begin{array}{rrrrrr} p-2 & -2 & -2 & -2 & -2 & -2 \\ -2 & p-2 & -2 & -2 & -2 & -2 \\ -2 & -2 & p-2 & -2 & -2 & -2 \\ -2 & -2 & -2 & p-2 & -2 & -2 \\ -2 & -2 & -2 & -2 & p-2 & -2 \\ -2 & -2 & -2 & -2 & -2 & p-2 \end{array}\right)$$
The diagonal is formed by the sums (with $q=(p-1)/2)$): $$d_{i,i}=\sum_{k=1}^qc_k^2=\sum_{k=1}^q(\zeta^k+\zeta^{-k})^2=2q-\frac{\zeta}{1+\zeta}-\frac{1}{1+\zeta}=p-2$$
The cyclic permutation $\sigma$ acts on the orbits (and rows): note $k(i)$ the power of $\sigma$ for the $i$th row of $D$, given by $k:(1,2,3,4,5,6)\to (6,1,4,2,3,5)$ as seen above.
The non-diagonal terms are: $$d_{i,j}=\sum_{r=1}^q c_{\sigma^{k(i)}(r)}c_{\sigma^{k(j)}(r)}=2\sum_{r=1}^q c_{\sigma^{k(i)}(r)+\sigma^{k(j)}(r)} +\sum_1^q c_{\sigma^{k(i)}(r)-\sigma^{k(j)}(r)}$$ with $2\cos a\cos b=\cos(a+b)+\cos(a-b)$. Since we permute all this $i$'s, we have:
$$d_{i,j}=2\sum_1^q c_i=2\sum_1^q (\zeta^i+\zeta^{-i})=-2$$
Then one computes (via SAGE) the determinant $\det M\cdot M^t=-p^{p-3}$ so $\det M=-\sqrt{p}^{p-3}$.
I found it rather laborious to determine the automorphisms of $G(L/K)$ and didn't found a smarter route that the following.
For $p\equiv 1$, we have the tower of Galois (real) fields: $$\mathbb{Q}=K\subset K(\sqrt{p})=F\subset F(c)=L\subset M=K(\zeta)\subset \mathbb{C}$$ of respective degrees $2/\frac{p-1}4/2$.
One can compute the minimal of $c=2\cos(2\pi/p)$ over different subfields of it's splitting field $L=F(c)$ and check what the automorphisms do to the generators $\sqrt{p}$ and $c=2\cos(2\pi/p)$ of $L$. For the case $p=13$, the only subfield is $F=K(\sqrt{p})$ since $[F:K]=6$. namely over $F$.
For $p=13$, the minimal polynomial of $c=2\cos(2\pi/13)$ over $K$ is calculated from $\Phi_{13}$ with the change of variables $Z=X+1/X$, which is: $$P_{\min,c}=T^6 + T^5 - 5T^4 - 4T^3 + 6T^2 + 3T - 1\in K[T]\;(1)$$ $$P_{\min,c}=\prod_{g\in G(L/K)} (T-g(2\cos(2\pi /p))=\prod_{k=1}^{k=\frac{p-1}{2}} (T-2\cos(2\pi k/p)\in L[T]$$ and over the Galois subextension $F$, with $m=(p-1)/4$ and $\tau:\sqrt{p}\mapsto -\sqrt{p}\in G(L/K)$ $$P_{\min,c}=(T^m+a_1(\sqrt{p})T^{m-1}+\cdots+a_m)(T^m+\tau(a_1)T^{m-1}+\cdots+\tau(a_m))\in F[T]$$ $$P_{\min,c}=\left(T^3+(a_1+b_1\sqrt{p})T^2+(a_2+b_2\sqrt{p})T+a_3+b_3\sqrt{p}\right)$$ $$\times \left(T^3+(a_1-b_1\sqrt{p})T^2+(a_2-b_2\sqrt{p})T+a_3-b_3\sqrt{p}\right)$$
so expanding and solving the constants of this last equation with $(1)$ gives a solution $a_1=b_1=-\frac 12$, $a_2=-1,b_2=0$ and $a_3=-\frac 32, b_3=-\frac 12$ so $P_{\min,c}$ factors as $P_1P_2$ over $F$ as: $$P_{\min,c}=\left(T^3+\frac 12(1-\sqrt{p})T^2-T-\frac 12(3-\sqrt{p})\right)\times \left(T^3+\frac 12(1+\sqrt{p})T^2-T-\frac 12(3+\sqrt{p})\right)$$
Each irreducible factor has a cyclic Galois group of order $3$, and the calculated discriminant, $p$, a square in $F$, confirms this. Each factor corresponds to the automorphisms sending respectively $\sqrt{p}$ to $-\sqrt{p}$ and to $\sqrt{p}$.
The automorphisms of $G(L/K)$ are given by sending $c=2\cos(2\pi /13)$ to $c_k=2\cos(2\pi k/p)$ for $k\in [1,\dots ,(p-1)/2]$, and $\sqrt{p}\mapsto \pm \sqrt{p}$. We need to compute them explicitely. The Chebyshev polynomials give relations $\cos nx=T_n(\cos x)$ and, since each $c_i$ is a root of a cubic factor $P_i$, this will give quadratic polynomials in $c$ and linear in $\sqrt{p}$.
For example, $2\cos3x=2T_3(\cos x)=(2\cos x)^{3}-3(2\cos x)$. Since $P_1(c)=0$, then $c^3=\frac 12(-1+\sqrt{p})c^2+c+\frac 12(3-\sqrt{p})$, so these relations together gives the automorphism sending $$c\mapsto c_3=2\cos(6\pi /13)=\frac12(\sqrt{p}-1)c^2 - 2c +\frac12(3-\sqrt{p})$$
In the same way, $c_{4}=2T_4(c)=c_{1}^{4}-4c_{1}^{2}+2$ gives $$c\mapsto c_4=2\cos(8\pi/13)=\frac12(1-\sqrt{p})c^2 + c + \sqrt{p} - 2$$
We check that $(T-c)(T-c_3)(T-c_4)$ is the factor $P_1$ of $P_{\min ,c}$ over $F$.
We have $2\cos 2x=2T_2(\cos x)=(2\cos x)^2 -2$ so we have an automorphism sending $$c=2\cos(2\pi /13)\mapsto c_2=2\cos(4\pi /13)=c^2-2$$
The other automorphisms are:
$$c\mapsto c_5=2\cos(10\pi/13)=-2c^2 + \frac 12(1+\sqrt{p})c +\frac12(5-\sqrt{p})$$
$$c\mapsto c_6=2\cos(12\pi/13)=c^2 -\frac 12 (1+\sqrt{p})c - 1$$
and $(T-c_2)(T-c_5)(T-c_6)$ give the factor $P_2$.
Now we can compute the $[g_ig_j(c)]$ matrix. Since $G(L/K)$ is abelian, the matrix $M(c)=[g_ig_j(c)]$ with $g_i\in G(L/K)$ is symmetric. It's determinant is the product of the eigenvalues of it's characteristic polynomial, so one computes $\det(T\cdot I-M(c))$.
$$\det(T\cdot I_6-M(c))=T^6 + (-\sqrt{13} + 1)T^5 + (-\sqrt{13} - 26)T^4 + (26\sqrt{13} - 26)T^3 + (26\sqrt{13} + 169)T^2 + (-169\sqrt{13} + 169)T - 169\sqrt{13}$$ The constant term $-169\sqrt{13}=-\sqrt{13}^{5}$ is the product of the eigenvalues, so is determinant of $M$. We can further factor this polynomial as $$=(T + 1) (T + \sqrt{p})^2 (T - \sqrt{p})^3$$ which shows the eigenvalues $\pm\sqrt{p}$ and $-1$ with multiplicites, so again $\det M(c)=-\sqrt{13}^5\in F$.
The case $p=5$ gives $\det M=-\sqrt{5}$.
