The quadratic subfield of $\mathbb{Q}(\zeta\_p)$ is given by $\mathbb{Q}(\sqrt{p^\*})$, where $p^\*$ is the choice of $\pm p$ which is $1$ mod $4$. By some elementary Galois theory, the cyclotomic polynomial $\Phi\_p = \frac{x^p-1}{x-1}$ factors into two irreducible polynomials of degree $\frac{p-1}{2}$ over this quadratic subfield, which one can describe as
$$
P\_{QR}(x) = \prod\_{k\ \text{QR}} (x-\zeta\_p^k)
$$
$$
P\_{QNR}(x) = \prod\_{k\ \text{QNR}} (x-\zeta\_p^k)
$$
where $k$ ranges over the (nonzero) quadratic residues and quadratic nonresidues modulo $p$ respectively.


Is there an explicit description for the coefficients of these polynomials in terms of $\sqrt{p^\*}$? The first coefficient, the one in front of $x^{\frac{p-3}{2}}$, is $\frac{1\mp\sqrt{p^\*}}{2}$ by the statement of Gauss sums.



Factoring cyclotomic polynomials over quadratic subfield
Some trivial observations. We have
$$P\_{QR}(1/x) x^{(p-1)/2} = \prod\_{QR} (1 - x \zeta^k),$$
$$P\_{QNR}(1/x) x^{(p-1)/2} = \prod\_{QNR} (1 - x \zeta^k),$$
which are easier to work with. On the other hand, 
if $p \ge 5$, the product of $\zeta^k\_p$ over quadratic residues is one,
and the product over non-residues is also one. Hence we can write
$$\begin{aligned}
P\_{QR}(1/x) x^{(p-1)/2} = & \ \prod\_{QR} ( 1 - x \zeta^k) = \prod\_{QR} ( \zeta^{-k} - x) \\
= & \ 
 (-1)^{(p-1)/2} \prod\_{QR} (x - \zeta^{-k}) \\
 = & \ \begin{cases} P\_{QR}(x), & p \equiv 1 \mod 4 \\
- P\_{QNR}(x), & p \equiv 3 \mod 4 \end{cases} \end{aligned}$$
because $(-1/p) = (-1)^{(p-1)/2}$. The same swapping occurs for $P\_{QNR}$.
Let's consider the highest powers of $P\_{QR}(x)$ and $P\_{QNR}(x)$,
or equivalently the lowest powers of $P\_{QR}(1/x) x^{(p-1)/2}$ and
$P\_{QNR}(1/x) x^{(p-1)/2}$. We have
$$\frac{P\_{QR}(1/x) x^{(p-1)/2}}{P\_{QNR}(1/x) x^{(p-1)/2}}
= \prod ( 1 - x \zeta^k)^{ \left( \frac{k}{p}\right)}.$$


As mentioned there is the Gauss sum:
$$\sum \left( \frac{k}{p}\right) \zeta^{k} = \sqrt{p^\*},$$
and similarly
$$\sum \left( \frac{k}{p}\right) \zeta^{nk} = \left( \frac{n}{p} \right) \sqrt{p^\*},$$
applying $[n] \in (\mathbb{Z}/p\mathbb{Z})^{\times} = \mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ to both sides; when $p|n$ we interpret the RHS as being zero, and this is still correct, although we won't actually care about terms this deep into the power series above.
We deduce that
$$- \log \left( \frac{P\_{QR}(1/x) x^{(p-1)/2}}{P\_{QNR}(1/x) x^{(p-1)/2}} \right) = \sum\_{n=1}^{\infty} \frac{x^n}{n} \sum \left( \frac{k}{p}\right) \zeta^{nk} 
= \sqrt{p^\*} \sum\_{n=1}^{\infty} \left( \frac{n}{p}\right) \frac{x^n}{n}.$$
Naturally
$$- \log \left( P\_{QR}(1/x) x^{(p-1)/2} P\_{QNR}(1/x) x^{(p-1)/2} \right)
= \log \left( \frac{1 - x^p}{1 - x} \right) = \log(1-x) + O(x^p) = - \sum \frac{x^n}{n},$$
and so (for example)
$$\log(P\_{QR}(1/x) x^{(p-1)/2}) = \frac{1}{2} \sum \frac{x^n}{n} \left(1 - \left( \frac{n}{p}\right) \sqrt{p^\*}\right) + O(x^p),$$
You can now formally expand this out to get the first few terms.
For example, the first non-zero term is
$$\frac{1 - \sqrt{p^\*}}{2},$$
and the second is
$$ \frac{3 + p^\* - 2 \sqrt{p^\*}\left(1 + \left( \frac{2}{p}\right) \right) }{8}$$
For example, if $p \equiv 3,5 \mod 8$ so $(2/p) = -1$, this is
$$\frac{3 + p^\*}{8}.$$
Note the conditions on $p$ ensure that this is an algebraic integer, as it has to be.
As you keep going, you get more and more terms involving the quadratic residues $(n/p)$ for small $n$,
and it becomes messier and messier, and dependent on $p$ modulo higher integers.
The third term, for example, is
$$\frac{15 - 9 \sqrt{p^\*} + 3 p^\* - p^\* \sqrt{p^\*} + 6(p^\* - \sqrt{p^\*})(2/p) - 8 \sqrt{p^\*} (3/p)}{48}.$$


