WLOG, let $\lambda = 1$ (rescale your problem appropriately, if necessary). Then, it is well-known consequence of Bernstein's inequality (e.g see theorem 3.1.1 of "High-dimensional Probability" book by R. Vershynin) that
$$ P(|\|X\|^2-n| \le n\epsilon) \ge 1 - 2e^{-Cn\min(\epsilon,\epsilon^2)},\;\forall \epsilon \ge 0. $$
Here, $C$ is a constant which is independent of $n$. In other words, $\|X\|^2$ as good (exponential / Gaussian) concentration around the value $n$.