Prove that $$( A) \quad \int _{-\pi}^{\pi} \sum_{n=2}^{\infty} \tan \frac{x}{2^n} \mathrm{d}x=\sum_{n=2}^{\infty}\int _{-\pi}^{\pi} \tan \frac{x}{2^n}\mathrm{d}x$$
We have $$ \forall n\ge 2, \quad \forall x\in (-\pi,\pi): \quad |\tan \frac{x}{2^n}|\le \tan \frac{\pi}{2^n}$$ and $$\lim_{n\to \infty}\frac{\tan \frac{\pi}{2^{n+1}}}{ \tan \frac{\pi}{2^n}}=\frac{1}{2}$$ We can understand the series $\sum _{n=2}^{\infty} \tan\frac{\pi}{2^n} $, which is convergent like has just been proved (by ratio test), as a function series with constant functions, thus it is uniformly convergent on $(-\pi,\pi)$ and so by the Weierstrass M-test the series $\sum_{n=2}^{\infty} \tan \frac{x}{2^n}$ is uniformly convergent on $(-\pi,\pi)$. Now we know that the equality (A) holds.
Your proof is perfectly fine!
Note that when using the M-test it is sufficient to show that $\sum_{n=2}^\infty \tan \frac{\pi}{2^n}$ is convergent. Considering this series as a uniformly convergent series of constant functions is not wrong, but also not necessary.
Finally, in this particular case you actually only need to prove that the function series converges pointwise (and thus yields a well-defined function), since both integrals are equal to zero by symmetry anyway. However, your proof works for other intervals of integration as well.