The easy way to solve this is to notice that $x_n = 2\cos(\pi 2^{-(n+1)})$, and first study the tower of fields $F_n = \Bbb Q(e^{2i\pi/2^{n+2}})$. It is easy to show that the Galois group of $F_n$ over $\Bbb Q$ is isomorphic to $(\Bbb Z / 2^{n+2} \Bbb Z)^*$ : pick a primitive $2^{n+2}$th root of unity $\zeta$. Then for $\alpha \in (\Bbb Z / 2^{n+2} \Bbb Z)^*$ there is a unique automorphism $\sigma_\alpha$ of $F_n$ sending $\zeta$ to $\zeta^\alpha$. $\sigma_{\alpha \beta} = \sigma_\alpha \circ \sigma_\beta$, and $\sigma_1 = id$.
Next, $E_n$ is the real subfield of $F_n$ : the subfield of $F_n$ fixed by complex conjugation. Since conjugation is $\sigma_{-1}$, the Galois group of $E_n$ is the quotient $(\Bbb Z / 2^{n+2} \Bbb Z)^*/\langle -1 \rangle$.
Finally, we need more on the structure of $(\Bbb Z / 2^{n+2} \Bbb Z)^*$. It turns out that for $n\ge 1$, $3$, of order $2^n$ and $-1$, of order $2$, generate it : they give an isomorphism $\Bbb Z/2^n\Bbb Z \times \{\pm 1\} \to (\Bbb Z / 2^{n+2} \Bbb Z)^* $. Since we are quotienting by $\{\pm 1\}$, the resulting group is isomorphic to $\Bbb Z/2^n\Bbb Z$
As a bonus, we get an explicit generator of $Aut(E)$ (the one corresponding to $ \pm 3$ in $(\Bbb Z/2^{n+2} \Bbb Z)^*/\langle -1 \rangle $): let $P(X) = X^3-3X$ and $Q(X) = X^2-2$ (so that $Q(x_{n+1}) = x_n$), and define $\sigma_n$ on $E_n$ by $\sigma_n(x_n) = P(x_n)$. We check that it is a compatible family of field automorphisms :
$P \circ Q = (X^2-2)^3-3(X^2-2) = X^6-6X^4+9X²-2 = (X^3-3X)^2-2 = Q \circ P$
So we have : $\sigma_0(0) = P(x_0) = 0$.
$\sigma_{n+1}(x_n) = \sigma_{n+1}(Q(x_{n+1})) = Q(P(x_{n+1})) = P(Q(x_{n+1})) = P(x_n) = \sigma_n(x_n)$ .
Thus they give field automorphisms and $\sigma_{n+1}|_{E_n} = \sigma_n$ so we can glue them together to define $\sigma$ on $E_\infty = \cup E_n$.
The fact that $\sigma$ is a generator of the Galois group of $E_n$ translates to the fact that $P^{\circ 2^n} = X \pmod {Q^{\circ n}}$, and $P^{\circ 2^{n-1}} = -X \pmod {Q^{\circ n}}$