TITLE: What are the products $\prod_{A\subset{\mathbb F}_p\colon |A|=n} \sum_{a\in A} \zeta^a$ equal to?
QUESTION [18 upvotes]: 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 ${\mathbb F}_p$.

REPLY [2 votes]: Only a partial answer:
I can prove that ${\cal P}_p(3)=\epsilon_p v_p^p$, with $\varepsilon_p^2=1$, for every integer $p\ge 3$ (not necessarily prime). The proof is too technical to present here.
The values of $v_p$ are given by
$$ v_p^6=\frac{9W_p}{(2^p-(-1)^p)^2(1-(-2)^p)}, $$
where 
$$ W_p=\prod_\zeta\{(1+\zeta)^p-(-1)^p\}; $$
except for the sign these $W_p$ are the numbers of sequence A096964 in OEIS 
(Wendt's determinant). 
Also $W_p=\det(M+(1-(-1)^p) I)$ where $M$ is the circulant matrix 
with first line $\binom{p}{k}$ for $0\le k\le p-1$. 
The sequence $\varepsilon_p$ appear to have a simple periodic structure, but 
this I have not proved.