Let $\alpha, \beta, \gamma$ be the reduced arguments of the complex numbers $a, b, c$. Since $abc \in \mathbb{R}$, it follows that $\sin(\alpha + \beta + \gamma) = 0$, i.e., $\alpha + \beta + \gamma = k\pi$, where $k \in \mathbb{Z}$. Therefore, we can write $\gamma = k\pi - \alpha - \beta$ for some integer $k$. Furthermore, from $a+b+c \in \mathbb{R}$ we have $\sin \alpha + \sin \beta + \sin \gamma = 0$, thus $\sin \alpha + \sin \beta = -\sin \gamma = -\sin(k\pi - \alpha - \beta)$, leading to $|\sin \alpha + \sin \beta| = |\sin(\alpha + \beta)|$, which is equivalent to:
$$\left| \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \right| = \left| \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} \right|.$$
If $\sin \frac{\alpha+\beta}{2} = 0$, then there is an integer $q$ such that $\frac{\alpha+\beta}{2} = q\pi$, and $\gamma = (k-2q)\pi$, hence $\sin \gamma = 0$, which leads to $c \in \mathbb{R}$.
If $\sin \frac{\alpha+\beta}{2} \neq 0$, then $|\cos \frac{\alpha+\beta}{2}| = |\cos \frac{\alpha-\beta}{2}|$, from which we have $\cos^2 \frac{\alpha+\beta}{2} = \cos^2 \frac{\alpha-\beta}{2}$, thus $\cos(\alpha + \beta) = \cos(\alpha - \beta)$, leading to $\sin \alpha \sin \beta = 0$, i.e., at least one of the numbers $a$ or $b$ is real.
Therefore, at least one of the numbers $a, b$, or $c$ is real. Let this number be $a$. From the hypothesis, we obtain that $b+c \in \mathbb{R}$, and since $a$ is nonzero, we also have $bc \in \mathbb{R}$. If $b$ is real, then $c$ will also be real, so $C_n$ is real for any $n \in \mathbb{N}$. If $b \in \mathbb{C} \setminus \mathbb{R}$, then $b$ and $c$ are the roots of a quadratic equation with real coefficients, thus $b = \bar{c}$. Consequently, $C_n = a^n + b^n + c^n = a^n + b^n + \bar{b}^n = a^n + b^n + \bar{c}^n \in \mathbb{R}$.

Alternative solution.
Let $|a| = |b| = |c| = r > 0$. For a complex number $z$ with modulus $r > 0$, we have $\bar{z} = \frac{r^2}{z}$. Since $A$ is a real number, we obtain:
$$a+b+c = \bar{a}+\bar{b}+\bar{c} = \frac{r^2}{a}+\frac{r^2}{b}+\frac{r^2}{c} = r^2 \cdot \frac{ab+bc+ca}{abc} \in \mathbb{R},$$
which implies $ab + bc + ca \in \mathbb{R}$.
For $n = 0$ we have $C_0 = 3$, and for $n = 1$ we have $C_1 = A \in \mathbb{R}$. Also, for $n = 2$ we have $C_2 = A^2 - 2(ab + bc + ca) \in \mathbb{R}$. For $n \ge 2$, we have:
$$C_{n+1} = A \cdot C_n - (ab + bc + ca) \cdot C_{n-1} + B \cdot C_{n-2}.$$
By induction, it now follows that $C_n \in \mathbb{R}$ for any $n \in \mathbb{N}$.