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.