Put $z^{3n} + \bar{z}^{3n} = f_n(z)$. As $f_n(z) = \overline{f_n(z)}$, we get $f_n(z) \in \mathbb{R}$, for all $z \in \mathbb{C}$. We should determine the set $M = \{z \in \mathbb{C} / f_n(z) \ge 0\}$. If $z \in M$, then $f_0(z) = z + \bar{z} \ge 0$, so $\text{Re}(z) \ge 0$.

For non-negative $z \in \mathbb{R}$, $f_n(z) = 2z^{3n} \ge 0$, for all non-negative integers $n$, implying $[0, \infty) \subset M$.

If $z = bi \in \mathbb{C}$, $b \in \mathbb{R}$, then $\bar{z} = -z$, so $f_n(z) = 0$, for $n \in \mathbb{N}$, so $\{bi / b \in \mathbb{R}\} \subset M$. As $f_n(z) = f_n(\bar{z})$, we have $z \in M \Leftrightarrow \bar{z} \in M$.

It remains to find $z \in M$ such that $\text{Re}(z) > 0$ and $\text{Im}(z) > 0$. Let $z = r(\cos t + i \sin t)$ with $r > 0$, $t \in (0, \frac{\pi}{2})$. Then $f_n(z) = 2r^{3n} \cos(3^n t)$, so $z \in M \Leftrightarrow \cos(3^n t) \ge 0$, for all $n \in \mathbb{N}$.

As $t \in (0, \frac{\pi}{2})$, there is $k \in \mathbb{N}$ such that $3^k t \le \frac{\pi}{2} < 3^{k+1}t$ (in fact $k = \lfloor \log_3 \frac{\pi}{2t} \rfloor$). If $3^k t = \frac{\pi}{2}$, then $\cos(3^n t) > 0$, for all $n \in \{0, 1, 2, \dots, k-1\}$ and $\cos(3^n t) = 0$, for all $n \ge k$, that is $z \in M$. If $3^k t < \frac{\pi}{2} < 3^{k+1}t$, then $\frac{\pi}{2} < 3^{k+1}t < \frac{3\pi}{2}$, implying $\cos(3^{k+1}t) < 0$, that is $z \notin M$.

In conclusion
$$M = [0, \infty) \cup \left\{ r \left( \cos \frac{\pi}{2 \cdot 3^k} \pm i \sin \frac{\pi}{2 \cdot 3^k} \right) / r > 0, k \in \mathbb{N} \right\}.$$