The result depends on the approximation properties of $\alpha$.

Of course one has to assume $\int_{S^1} f(z)dz=0$. A rotation by $\alpha$ has the effect that the $k$-th Fourier coefficient of $f$ is multiplied by $\exp(2\pi i \cdot k \alpha)$. Hence, the $k$-th Fourier coefficient (for $k \neq 0$) of $T_n(f)$ is just
$$ \frac1{\sqrt{n}} \sum_{l=1}^n \exp(2\pi i \cdot k l \cdot\alpha) \cdot \hat f(k) =\frac1{\sqrt{n}} \cdot \exp(2\pi i \cdot k \cdot\alpha) \cdot \frac{1 - \exp(2\pi i \cdot k n \cdot\alpha)}{1 - \exp(2\pi i \cdot k \cdot\alpha)} \cdot \hat f(k).$$

If $\alpha$ is algebraic (or diophantine generic), then
$|\alpha - p/k| \geq C/k^M$ for some constants $C$ and $M$. Hence,
$|1 - \exp(2\pi i \cdot k \cdot\alpha)|^{-1}$ grows at most like a polynomial in $k$. If $f$ is smooth, then $k \mapsto \hat f(k)$ decays rapidly, so that $k \mapsto |1 - \exp(2\pi i \cdot k \cdot\alpha)|^{-1} \hat f(k)$ is still in $\ell^1(\mathbb Z)$.
This altogether implies that $T_n(f)$ converges to zero uniformly on $S^1$.

On the other side, if $\alpha$ is some well-chosen Liouville number and $f$ some special constructed smooth function, then I believe that $T_n(f)$ need not converge pointwise (or uniformly).