Solution:

Since $\tan$ has period $180^\circ,$ only $2^n\theta \bmod 180^\circ$ matters: the tangent is positive on $(0^\circ, 90^\circ)$ and negative on $(90^\circ, 180^\circ).$ Suppose $\theta$ satisfies the condition, and let $\theta'$ be the reduction of $8\theta$ modulo $180^\circ;$ since $\tan(8\theta) \gt 0,$ we have $\theta' \in (0^\circ, 90^\circ).$ For every $n,$ $2^n\theta' \equiv 2^{n+3}\theta \pmod{180^\circ},$ and the sign pattern for the exponents $n + 3$ is the same as for $n,$ so $\theta'$ also satisfies the condition. By uniqueness, $\theta' = \theta,$ so $7\theta \equiv 0 \pmod{180^\circ}$ and $\theta = \frac{180k}{7}$ degrees for some $k \in \{1, 2, 3\}.$

Test each: for $k = 1,$ $2\theta = \frac{360^\circ}{7} \approx 51.4^\circ$ has positive tangent — fails. For $k = 2,$ $4\theta = \frac{1440^\circ}{7} \equiv \frac{180^\circ}{7} \approx 25.7^\circ$ has positive tangent — fails. For $k = 3,$ $\theta = \frac{540^\circ}{7} \approx 77.1^\circ:$ then $2\theta \approx 154.3^\circ$ and $4\theta \equiv 128.6^\circ$ are both in $(90^\circ, 180^\circ),$ and $8\theta \equiv \theta,$ so the pattern positive, negative, negative repeats forever.

Thus $\theta = \frac{540}{7}$ degrees, and $p + q = 540 + 7 = 547.$