Here is my attempt. This is motivated by Jack D'Aurizio's trick os using Fourier series expansion of $\log\cos(x)$. I need some more eyes to go through and see if I made any mistakes. Please feel free to edit/fix/add if you find it helpful.
\begin{eqnarray*}
\log \cos(x) &=& \log \frac{1}{2} -\sum_{m\ge 1}^{\infty}{\frac{(-1)^{m} \cos(2 m x)}{m}}
\end{eqnarray*}
\begin{eqnarray*}
\sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\cos \frac{2\pi p k}{n}} &=& \begin{cases} -\frac{1}{2}, & mod(n,p) \ne 0 \\ \left \lfloor \frac{n}{2} \right\rfloor, & mod(n,p) =0 \end{cases}
\end{eqnarray*}
Generically,
\begin{eqnarray*}
\sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\cos \frac{2\pi p k}{n}} &=& -\frac{1}{2} \left[1-\frac{\sin\left(\pi p \frac{1+2\left\lfloor \frac{n}{2}\right\rfloor}{n}\right)}{\sin \frac{\pi p}{n}} \right] \\
&=& \begin{cases} \frac{n-1}{2} , & mod(p,n) = 0, n, odd \\ -\frac{1}{2} , & mod(p,n) \ne 0, n, odd \\ -\frac{1}{2} \left[1-(-1)^{p}\right], & n, even\end{cases} \\
&=& \begin{cases} \left\lfloor\frac{n}{2}\right\rfloor , & mod(p,n) = 0, n, odd \\ -\frac{1}{2} , & mod(p,n) \ne 0, n, odd \\ -1, & n, even, p, odd \\ 0, & n, even, p, even \end{cases}
\end{eqnarray*}
Note that,
\begin{eqnarray*}
\frac{1+2\left\lfloor \frac{n}{2}\right\rfloor}{n} &=& \begin{cases} 1, & n, odd \\ 1+\frac{1}{n}, & n, even \end{cases} \\
\sin\left(\pi p \frac{1+2\left\lfloor \frac{n}{2}\right\rfloor}{n}\right) &=& \begin{cases} 0, & n, odd \\ (-1)^{p}\sin \frac{\pi p}{n}, & n, even \end{cases} \\
\end{eqnarray*}
For $x=\frac{\pi p k}{n}$,
\begin{eqnarray*}
\sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor} \sum_{m\ge 1}^{\infty}{\frac{(-1)^{m} \cos \left(\frac{2 \pi p k m}{n}\right)}{m}} &=& \sum_{m\ge 1}^{\infty} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\frac{(-1)^{m} \cos \left(\frac{2 \pi p k m}{n}\right)}{m}} \\
&=& \sum_{\underset{m\ge 1}{m \ne \frac{n}{(n,p)} \mathbb{Z} }}^{\infty} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\frac{(-1)^{m} \cos \left(\frac{2 \pi p k m}{n}\right)}{m}} + \sum_{\underset{m\ge 1}{m = \frac{n}{(n,p)} \mathbb{Z} }}^{\infty} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\frac{(-1)^{m} \cos \left(\frac{2 \pi p k m}{n}\right)}{m}} \\
&=& \sum_{\underset{m\ge 1}{m \ne \frac{n}{(n,p)} \mathbb{Z} }}^{\infty} \frac{(-1)^{m}}{m} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{ \cos \left(\frac{2 \pi p k m}{n}\right)} + \sum_{\underset{m\ge 1}{m = \frac{n}{(n,p)} \mathbb{Z} }}^{\infty} \frac{(-1)^{m}}{m} \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{ \cos \left(\frac{2 \pi p k m}{n}\right)} \\
&=& \sum_{\underset{m\ge 1}{m \ne \frac{n}{(n,p)} \mathbb{Z} }}^{\infty} \frac{(-1)^{m}}{m} \left(-\frac{1}{2}\right) + \sum_{\underset{m\ge 1}{m = \frac{n}{(n,p)} \mathbb{Z} }}^{\infty} \frac{(-1)^{m}}{m} \left\lfloor \frac{n}{2}\right\rfloor \\
&=& \sum_{m\ne N_{p}}^{\infty} \frac{(-1)^{m}}{m} \left(-\frac{1}{2}\right) + \sum_{m= i N_{p},\forall i}^{\infty} \frac{(-1)^{m}}{m} \left\lfloor \frac{n}{2}\right\rfloor \\
&=& \sum_{m\ne N_{p}}^{\infty} \frac{(-1)^{m}}{m} \left(-\frac{1}{2}\right) + \sum_{i=1}^{\infty} \frac{(-1)^{ i N_{p}}}{ i N_{p}} \left\lfloor \frac{n}{2}\right\rfloor \\
&=& -\frac{1}{2} \sum_{m=1}^{\infty} \frac{(-1)^{m}}{m} + \frac{1}{2} \sum_{i=1}^{\infty} \frac{(-1)^{ i N_{p}}}{ i N_{p}} \left(1+2 \left\lfloor \frac{n}{2}\right\rfloor\right)
\end{eqnarray*}
Here $N_p \triangleq \frac{n}{gcd(n,p)}$.
For even $N_p$, the harmonic series diverges. The only interesting case is $N_p$ being odd. In which case,'
\begin{eqnarray*}
&=&-\frac{1}{2}\sum_{m=1}^{\infty} \frac{(-1)^{m}}{m} + \frac{1}{2} \sum_{m=1}^{\infty} \frac{(-1)^{ m N_{p}}}{ m N_{p}} \left(1+2 \left\lfloor \frac{n}{2}\right\rfloor\right) \\
&=&\frac{1}{2} \sum_{m=1}^{\infty} \frac{(-1)^{ m}}{ i} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \\
&=&\frac{1}{2} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \sum_{m=1}^{\infty} \frac{(-1)^{ m}}{ i} \\
&=&\frac{1}{2} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \log(2).
\end{eqnarray*}
\begin{eqnarray*}
\sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\log \cos\left(\frac{\pi p k}{n}\right)} &=& \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor} \log \frac{1}{2} - \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor} \sum_{m\ge 1}^{\infty}{\frac{(-1)^{m} \cos\left(\frac{2\pi p k m}{n}\right)}{m}} \\
&=& \left\lfloor \frac{n}{2}\right\rfloor \log \frac{1}{2} - \sum_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor} \sum_{m\ge 1}^{\infty}{\frac{(-1)^{m} \cos\left(\frac{2\pi p k m}{n}\right)}{m}} \\
&=& \left\lfloor \frac{n}{2}\right\rfloor \log \frac{1}{2} - \frac{1}{2} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \log(2) \\
&=& \left\lfloor \frac{n}{2}\right\rfloor \log \frac{1}{2} + \frac{1}{2} \left(\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \log \frac{1}{2} \\
&=& \frac{1}{2} \left(2 \left\lfloor \frac{n}{2}\right\rfloor+\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{N_p} -1\right) \log \frac{1}{2} \\
&=& \log \left[\frac{1}{2^{ \left( \left\lfloor \frac{n}{2}\right\rfloor+\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{2 N_p} -\frac{1}{2}\right)}} \right]
\end{eqnarray*}
Looking at the exponent,
\begin{eqnarray*}
\prod_{k=1}^{\left\lfloor \frac{n}{2}\right\rfloor}{\cos\left(\frac{\pi p k}{n}\right)}
&=& 2^{- \left( \left\lfloor \frac{n}{2}\right\rfloor+\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{2 N_p} -\frac{1}{2}\right)} \\
&=& 2^{- \left( \left\lfloor \frac{n}{2}\right\rfloor+\frac{1+2 \left\lfloor \frac{n}{2}\right\rfloor}{2 \frac{n}{gcd(n,p)}} -\frac{1}{2}\right)}.
\end{eqnarray*}
A special case is when $p=1$ and $n$, $N_p=n/gcd(n,p)=n$, it becomes $2^{-n}$, the well known formula linked above.
