Well, in your case notice that $\|X\|^2>0$ and so $\mu\triangleq \mathbb{E}[\|X\|^2]>0$.  Thus, for any $\lambda \in \left(0,\mu\right)$ the Cantelli Inequality gives
$$
\Pr(X\ge\lambda)
\ge 1 - \frac{\sigma^2}{\sigma^2 + \lambda^2},
$$
where $\sigma\triangleq \mathbb{E}\left[\left(\|X\|^2 - \mu\right)^2\right]$.