Solution:
Note that each of the following is equivalent to the next.

$\cdot\ \operatorname{sgn}(\sin (2^{n})) = +1$

$\cdot\ 0 < 2^{n} \bmod 2\pi < \pi$

$\cdot\ 0 < \frac{2^{n}}{\pi} \bmod 2 < 1$

The $n$th digit after the decimal point in the binary representation of $\frac{1}{\pi}$ is $0$.

Similarly, $\operatorname{sgn}(\sin (2^{n})) = -1$ if and only if the $n$th digit after the decimal point in the binary representation of $\frac{1}{\pi}$ is $1$. In particular, if $a_{n}$ is the $n$th digit, then $\operatorname{sgn}(\sin (2^{n})) = 1 - 2a_{n}$.

Thus, the desired sum is
$$\sum_{n = 1}^{\infty}\frac{\operatorname{sgn}(\sin(2^{n}))}{2^{n}} = \sum_{n = 1}^{\infty}\frac{1 - 2a_{n}}{2^{n}} = \left(\sum_{n = 1}^{\infty}\frac{1}{2^{n}}\right) - 2\left(\sum_{n = 1}^{\infty}\frac{a_{n}}{2^{n}}\right) = \left[1 - \frac{2}{\pi}\right].$$