I don't think that $p$ being prime makes any difference. None of what I am about to write is news but it is worth saying.

Let $f(x)=\sum_{j=0}^nc_jx^j. $ One might choose to have $c_n=1$ or $c_0=1$ but it is perhaps nicer not to. Then $\prod_{i=0}^{n-1} f(\omega^ix)=\sum_{j=0}^nC_j(x^n)^j$.
One can say that 

 - $C_j$ is a polynomial of degree $n$ in the coefficients $c_0,\cdots,c_n$ where each term has total degree $n$ 
 - $C_j$ has a term $(-1)^{nj}(c_j)^n$ and no term $c_j^{n-1}$.
 - $C_{n-j}$ is $C_j$ with $c_k$ replaced by $c_{n-k}$

The $c_i$ are symmetric polynomials of the $n$ roots $\alpha_i$ of $f$ ( for $i<n$, $c_n$ is a constant). The $\alpha_i$ can be thought of as formal variables. Then $c_0,\cdots,c_{n-1}$ are also a basis for the ring of all symmetric polynomials ($\frac{1}{c_n}$ times the a usual basis) . There are other bases for this ring such as $\sigma_k=\sum_{i=1}^n\alpha_i^k$ and the sum of all $\binom{n}{k}$ products of $k$ roots. Transforming between these bases (mre generally, expressing a given symmetric polynomial in terms of them) is a major topic of invariant theory.

In this case we want to express certain symmetric polynomials in the $\alpha_i^n$ (scaled by a power of $c_n$). This must be a well known case. At any rate here are some results:

For $n=4,$

$$C_0=c_0^4$$

$$C_1=(4c_0^3c_4-2c_0^2c_2^2)-(4c_0^2c_3)c_1+(4c_0c_2)c_1^2-c_1^4$$


$$C_2=(6c_0^2c_4^2-8c_0c_1c_3c_4+2c_1^2c_3^2)+(4c_0c_3^2+4c_1^2c_4)c_2+(-4c_0c_4-4c_1c_3)c_2^2+c_2^4$$


$$C_3=(4c_4^3c_0-2c_4^2c_2^2)-(4c_4^2c_1)c_3+4(c_4c_2)c_3^2-c_3^4$$


$$C_4=c_4^4$$

while for $n=5$ we have the following (with the others obtainable by symmetry)

$$C_0=c_0^5$$


$$C_1=5c_0^4c_5-(5c_0^3c_4-5c_0^2c_2^2)c_1+(5c_0^2c_3)c_1^2-5c_0^3c_2c_3-(5c_0c_2)c_1^3+c_1^5$$


$${\small C_2=(10c_0^3c_5^2-15c_0^2c_1c_4c_5+5c_0c_1^2c_4^2+5c_0^2c_3^2c_4+10c_0c_1^2c_3c_5-5c_0c_1c_3^3-5c_1^3c_3c_4)+(5c_0^2c_4^2-15c_0^2c_3c_5+5c_1^2c_3^2-5c_0c_1c_3c_4-5c_1^3c_5)c_2}$$ $${\small +(5c_1^2c_4+10c_0c_1c_5+5c_0c_3^2)c_2^2+(-5c_0c_4-5c_1c_3)c_2^3+c_2^5}$$