A typical trick is to use Chebyshev's inequality with the function $\exp(\alpha x)$ to obtain $P(\|X\|_1>t)\leq E(\exp(\alpha\|X\|_1))/\exp(\alpha t).$ Since the 1-norm is a sum and the coordinates are independent, the right hand side equals $\exp(-\alpha t)\ E(\exp(\alpha |Z|))^d=\exp(-\alpha t)\left(\exp(\alpha^2/2) (1+\mbox{erf}(\alpha/\sqrt{2})) \right)^d.$ Here $Z$ is a one dimensional standard normal random variable.
You can now try to optimize over $\alpha$. Substituting the cheap upper bound $\mbox{erf}(\alpha/\sqrt{2})\leq 1$, and then optimizing gives $\alpha=t/d$. Plugging this in we get
$ P(\|X\|_1>t)\leq 2^d \exp(-t^2/2d).$
I know you could do better, but maybe it will suffice for your purposes.