One can compute the probability of each dyadic event $[k\leqslant 2^nY_n\lt k+1]$, as Michael suggested in comments, or one can use characteristic functions.
For every real number $t$, let $\varphi_n(t)=\mathrm E(\mathrm e^{\mathrm itY_n})$. Note that $\mathrm E(\mathrm e^{\mathrm itX_1})=\cos(t)$. By independence, $ \varphi_n(t)=\prod\limits_{k=1}^{n}\mathrm E(\mathrm e^{\mathrm itX_1/2^k})=\prod\limits_{k=1}^{n}\cos(t/2^k). $ Here comes a trick: multiply the product on the RHS by $\sin(t/2^n)$ and use recursively from $k=n$ to $k=1$ the relation $2\sin(t/2^k)\cos(t/2^k)=\sin(t/2^{k-1})$. For every $t$ which is not a multiple of $2^n\pi$, this yields $ \varphi_n(t)=\frac{\sin(t)}{2^n\sin(t/2^n)}. $ When $n\to\infty$, $2^n\sin(t/2^n)\to t$ hence, for every $t\ne0$, $\varphi_n(t)\to\varphi(t)=\sin(t)/t$. Since $\varphi(t)=\mathrm E(\mathrm e^{\mathrm itU})$ with $U$ uniform on $(-1,1)$, this proves that $Y_n\to U$ in distribution.
Note: A stronger result holds: $Y_n\to Y$ almost surely, where $Y=\sum\limits_{n=1}^{+\infty}\frac{X_n}{2^n}$ is uniform on $(-1,1)$.