I want to prove this without using any of the properties about the field of algebraic numbers (specifically that it is one). Essentially I just want to find a polynomial for which $\cos\frac{2\pi}{n}$ is a root.
I know roots of unity and De Moivre's theorem is clearly going to be important here but I just can't see how to actually construct the polynomial from these facts.
We know $\cos(2\pi/n)+i\sin(2\pi/n)$ is algebraic ($n$th root of unity). And the complex conjugate of algebraic is algebraic (in fact it's just another $n$th root of unity), so add it to its conjugate because also the sum of two algebraics is algebraic. Then divide by two.
Let $\theta=\frac{2\pi}{n}$. By De Moivre's formula we have $$ (\cos \theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta = 1 $$
Expand the left-hand side using the binomial theorem!
Every other term of the expansion contains a $\pm i$ and an odd power of $\sin\theta$; we know these terms cancel out because the imaginary part of the right-hand side is zero.
If we remove all those, what is left is real terms with only even powers of $\sin \theta$. Substitute $(\sin\theta)^{2n} = (1-\cos^2\theta)^n$. Now you have a polynomial identity with integer coefficients in $\cos \theta$ -- in other words, an integer polynomial with $\cos\theta$ as root. Thus $\cos\theta$ is algebraic.
[This is actually one way to derive the Chebyshev polynomials].
Are you familiar with the Chebyshev polynomials $T_n$ (Wikipedia link)? One property is that $$T_n(\cos(\theta))=\cos(n\theta)$$ (see this section). Thus $$T_n(\cos(\tfrac{2\pi}{n}))=\cos(2\pi)=1$$ and therefore an example of a polynomial with $\cos(\frac{2\pi}{n})$ as a root is $T_n-1$.
Euler's Formula implies $$ \left[\cos\left(\frac{2\pi}n\right)+i\sin\left(\frac{2\pi}n\right)\right]^n=1\tag{1} $$ The Binomial Theorem says $$ \begin{align} 1 &=\sum_{k=0}^ni^k\binom{n}{k}\cos^{n-k}\!\left(\frac{2\pi}n\right)\sin^k\!\left(\frac{2\pi}n\right)\\ &=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k}\!\left(\frac{2\pi}n\right)\sin^{2k}\!\left(\frac{2\pi}n\right)\\ &+i\sum_{k=0}^{\lfloor(n-1)/2\rfloor}(-1)^k\binom{n}{2k+1}\cos^{n-2k-1}\!\left(\frac{2\pi}n\right)\sin^{2k+1}\!\left(\frac{2\pi}n\right)\tag{2} \end{align} $$ Looking at the real part of $(2)$ yields $$ \begin{align} 1 &=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k}\!\left(\frac{2\pi}n\right)\sin^{2k}\!\left(\frac{2\pi}n\right)\\ &=\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{2k}\cos^{n-2k}\!\left(\frac{2\pi}n\right)\left[1-\cos^2\!\left(\frac{2\pi}n\right)\right]^k\\ &=\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^m\cos^{n-2m}\!\left(\frac{2\pi}n\right)\sum_{k=m}^{\lfloor n/2\rfloor}\binom{n}{2k}\binom{k}{m}\tag{3} \end{align} $$ Equation $(3)$ gives a polynomial equation whose root is $\cos\!\left(\frac{2\pi}n\right)$. Furthermore, the coefficient of $\cos^n\!\left(\frac{2\pi}n\right)$ is $2^{n-1}$, which implies that $2\cos\!\left(\frac{2\pi}n\right)$ is an algebraic integer. This polynomial also has a root of $1$, so it is reducible.
Example: Using $n=7$ in $(3)$, we have that $\cos\!\left(\frac{2\pi}7\right)$ satisfies $$ 64x^7-112x^5+56x^3-7x-1=0\tag{4} $$ Note that $1$ is also a root of $(4)$.
The fourier transform together with higher frequencies are polynomials in lower frequencies should be enough to write any of them as radicals of polynomials in the other frequencies. And of course that radicals of polynomials are algebraic.
An idea which generalizes to other contexts aswell but uses a little machinery from algebra. In your particular case, probably just using binomial expansion is better.
We know that $\cos(2\pi/n)=(e^{2\pi i/n}+e^{-2\pi i /n})/2$ and both $e^{2\pi i/n}$ and $e^{-2\pi i/n}$ are complex numbers which are roots of $x^n-1$, so these are algebraic integers. It is also clear that $x^n-1=(x-1)(1+x+...+x^{n-1})$ which implies (using Eisenstein maybe) that $1+x+...+x^{n-1}$ is the irreducible polynomial. Using the first isomorphism theorem:
$$\mathbb{Z}[x]/(1+x+...+x^{n-1})\approx Im(eval(e^{2\pi i/n}))\subseteq\mathbb{Z}[e^{2\pi i/n}]$$
Actually the containment is an equality (and this follows from the fact we have a monic irreducible polynomial). This is the case because $\mathbb{Z}[e^{2\pi i/n}]$ has a natural basis given by $\{1,\alpha,...\alpha^{n-2}\}$. Indeed, these generate any element because we may divide by $1+\alpha+...+\alpha^{n-1}$ in order to lower the degree. Also, they are linearly independent, otherwise it would contradict the minimality of degree of $1+x+...+x^{n-1}$ the irreducible annihilator.
Now comes the nice trick. Let me call $\beta=\alpha^{n-1}$. We may consider the multiplication ring homomorphism given by multiplication $m_{\alpha+\beta}(x)=(\alpha+\beta)x$. This is of course linear transformation in our module and we may represent it as a matrix in the basis $\{1,\alpha,...,\alpha^{n-1}\}$. By Cayley-Hamilton, which holds over any module, we find that $p(\lambda)=\det(M-\lambda I)$ annihilates $m_{\alpha+\beta}$, that is $p(m_{\alpha+\beta})(x)=0$ for any $x\in \mathbb{Z}[\alpha]$. Taking $x=1$ yields that $p(\alpha+\beta)=0$.
This teaches us that there is a monic polynomial that annihilates $\alpha+\beta$, explicitly something like $x^m+a_{m-1}x^m+...a_1x+a_o$. Now it is clear that:
$$\tilde{p}(x)=2^mx^m+2^{m-1}a_{m-1}x^m+...2a_1 x+a_o$$
is such that $\tilde{p}(\cos(2\pi/n))=\tilde{p}((\alpha+\beta)/2)=p(\alpha+\beta)=0$.
