This is a somewhat more explicit version of a question I have recently asked.

Let $p$ be an odd prime, and write $\zeta:=\exp(2\pi i/p)$ (any other primitive $p$th root of unity will do as well). For integer $n\in[0,p]$, the products $$ {\mathcal P}_p(n) := \prod_{\substack{A\subseteq{\mathbb F}_p \\ |A|=n}}\ \sum_{a\in A} \zeta^a $$ are rational integers; can they be found "explicitly"?

It is immediately seen that ${\mathcal P}_p(0)=0$ and ${\mathcal P}_p(1)=1$, and I can show that ${\mathcal P}_p(2)=\left(\frac2p\right)$ (it is not difficult to see that ${\mathcal P}_p(2)=\pm 1$, but to determine the sign is trickier). Also, we have ${\mathcal P}_p(p-n)=(-1)^{\binom{p}{n}}{\mathcal P}_p(n)$, so that ${\mathcal P}_p(p)=0$, ${\mathcal P}_p(p-1)=-1$, and ${\mathcal P}_p(p-2)=\pm 1$.

$\quad$ What are the values of ${\mathcal P}_p(n)$ for $n\in[3,p-3]$?

Some numerical data (thanks to Talmon Silver for the programming):

$\quad {\mathcal P}_5(3)=-1$

$\quad {\mathcal P}_7(3)=-2^7$

$\quad {\mathcal P}_{11}(3)=23^{11}$

$\quad {\mathcal P}_{13}(3)=159^{13}$

$\quad {\mathcal P}_{17}(3)=-24617^{17}$

$\quad {\mathcal P}_{19}(3)=-611009^{19}$

$\quad {\mathcal P}_{23}(3)=1265401351^{23}$

$\quad$ If finding the individual values ${\mathcal P}_p(n)$ is difficult, can we at least find explicitly the product $$ {\mathcal P}_p(1){\mathcal P}_p(2)\dotsb{\mathcal P}_p(p-2){\mathcal P}_p(p-1) = \prod_{\varnothing\ne A\subsetneq{\mathbb F}_p} \sum_{a\in A} \zeta^a \ ?$$

Denoting this product by ${\mathcal P}_p$,

$\quad {\mathcal P}_3=-1$

$\quad {\mathcal P}_5=-1$

$\quad {\mathcal P}_7=-2^{14}$

$\quad {\mathcal P}_{11}=-(3\cdot 23^4 \cdot 67\cdot 89)^{22}$

$\quad {\mathcal P}_{13}=-(3^{12}\cdot 5\cdot 53^6 \cdot 79^4\cdot 131^2 \cdot 157^2 \cdot 313\cdot 547\cdot 599\cdot 911)^{26}$

The problem can be restated in a purely combinatorial way, as hinted to in Ofir's comment below. Write $N:=\binom pn$, let $A_1,\dotsc,A_N$ be all the $n$-element subsets of ${\mathbb F}_p$, and for $z\in{\mathbb F}_p$ denote by $r_n(z)$ the number of representations $z=a_1+\dotsb+ a_N$ with $a_1\in A_1,\dotsc, a_N\in A_N$. We have then ${\mathcal P}_p(n)=\sum_{z\in{\mathbb F}_p}r_n(z)\zeta^z$, and from the fact that ${\mathcal P}_p(n)$ is an integer, it follows that $r_n(z)$ are actually equal to each other for all $z\in{\mathbb F}_p\setminus\{0\}$; as a result, we have, say, ${\mathcal P}_p(n)=r_n(0)-r_n(1)$. On the other hand, $$ r_n(0)+(p-1)r_n(1) = \sum_{z\in{\mathbb F}_p} r_n(z) = |A_1|\dotsb|A_N| = n^{\binom pn}. $$ This yields $$ {\mathcal P}_p(n) = \frac1{p-1} \left( p\,r_n(0)-n^{\binom pn}\right). $$ Thus, the problem boils down to finding $r_n(0)$, the number of all zero-sum $N$-tuples $(a_1,\dotsc,a_N)$ with the components $a_i$ representing each of the $n$-element subsets of ${\mathbb F}_p$.