We show that $n = 7$. Note that
$$\cos \pi = -1, \cos \frac{\pi}{2} = 0, \cos \frac{\pi}{3} = \frac{1}{2}, \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}.$$
Further, it follows from $0 = \cos \frac{3\pi}{5} + \cos \frac{2\pi}{5}$ that $x_5 = \cos \frac{\pi}{5}$ is a root of the equation $0 = 4x^3 - 3x + 2x^2 - 1 = (x+1)(4x^2 - 2x - 1)$, i.e. $x_5 = \frac{1+\sqrt{5}}{4}$.
It remains to prove that $x_7 = \cos \frac{\pi}{7}$ can not be expressed in the form $p + \sqrt{q} + \sqrt[3]{r}$, where $p, q$ and $r$ are rational numbers. Since $0 = \cos \frac{4\pi}{7} + \cos \frac{3\pi}{7}$ we have that $x_7$ is a root of
$$0 = 2(2x^2 - 1)^2 - 1 + 4x^3 - 3x = (x + 1)(8x^3 - 4x^2 - 4x + 1),$$
i.e. $x_7$ is a zero of $P(x) = 8x^3 - 4x^2 - 4x + 1$. Suppose that $x_7 = p + \sqrt{q} + \sqrt[3]{r}$, where $q \ge 0, p, r \in \mathbb{Q}$. Now $x_7$ is a zero of $Q(x) = (x - p - \sqrt{q})^3 - r$ with coefficients of the form $a + b\sqrt{q}$, $a, b \in \mathbb{Q}$, i.e. from $\mathbb{Q}[\sqrt{q}]$. Since the coefficient of $x$ is nonnegative we have that $P \ne 8Q$. Thus $P = 8Q + R$, where $R$ is a

polynomial of degree 1 or 2 with coefficients from $\mathbb{Q}[\sqrt{q}]$ and $R(x_7) = 0$. If $\deg R = 1$, then $x_7 \in \mathbb{Q}[\sqrt{q}]$. If $\deg R = 2$ and $R$ does not divide $P$, then $x_7$ is a zero of the remainder of $P$ divided by $R$, i.e. again $x_7 \in \mathbb{Q}[q]$. If $\deg R = 2$ and $R$ divides $P$ then the zero of $\frac{P}{R}$, which is from $\mathbb{Q}[q]$ is also zero of $P$. Thus, $P$ has a zero from $\mathbb{Q}[q]$. It is a zero of a polynomial of degree 1 or 2 with rational coefficients. It follows as above that $P$ has rational zero. Direct verification shows that none of the numbers $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}$ (which are all possible rational zeroes of $P$) is a zero of $P$, a contradiction.