This is indeed true. Assume WLOG that the matrices are diagonal. A useful way to handle the uniform measure on the sphere is that if you let $X_1,\ldots,X_n$ be iid (complex) Gaussians, then the vector $$\frac{1}{\sqrt{\sum_j |X_j|^2}}(X_1,\ldots,X_n)$$ is uniform on the sphere.
Standard concentration results show that, say, $$\frac{3}{4} \leq \frac{\sum_j |X_j|^2}{n} \leq \frac{4}{3} $$
with probability at least $1 - e^{-\Omega(n)}$.
Then you have \begin{align*} \mathbb{P}(|\langle T_n v_n, v_n\rangle| \geq\epsilon) &\leq \mathbb{P}\left(\left|\sum_j |X_j|^2 \frac{\lambda_j}{n} \right| \geq \frac{3}{4}\epsilon\right) + e^{-\Omega(n)} \\ &\leq \mathbb{P}\left(\left|\sum_j (|X_j|^2 -1)\lambda_j \right| \geq \frac{n}{4}\epsilon\right) + e^{-\Omega(n)} \\ &\leq e^{-\Omega(n)} \end{align*} where the second inequality is using the fact that $\left|\sum_{j} \lambda_j / n\right| \leq \epsilon/2$ and the last inequality is using a Chernoff bound, since $(|X_j|^2 - 1)\lambda_j$ are independent mean zero random variables with an exponential moment (uniformly bounded over $j$, since $|\lambda_j| =1$ for all $j$).
