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