Since you're trying to bound the sum of zero-mean i.i.d. RVs, I would recommend you try to develop a Chernoff bound: $$\Pr(X^TY>\epsilon)\leq \inf_{s\geq 0}(e^{-s\epsilon }(Ee^{sZ})^m)$$ where $Z=X_1Y_1$ is distributed according to a Normal Product distribution. I haven't carried out the calculation in full but I believe the moment generating function $Ee^{sZ}$ can be computed in close form using the expression (6) for $K_0$ found here.
As to tightness of the bound, notice that $$\Pr(X^TY>\epsilon)=\Pr(\sum_{i=1}^m\hat{Z}_i>m\epsilon)$$ where the $\hat{Z}_i$ are i.i.d. and each one is the product of two independent standard ($\mathcal{N}(0,1)$) Gaussian RVs. It is a standard Large Deviations result that such probability goes to zero exponentially fast as $m\to\infty$ for every constant $\epsilon>0$. I am 99% sure that the Chernoff bound always yields the correct exponential rate (but not the correct coefficient of the leading exponent).
Since you're trying to bound the sum of zero-mean i.i.d. RVs, I would recommend you try to develop a Chernoff bound: $$\Pr(X^TY>\epsilon)\leq \inf_{s\geq 0}(e^{-s\epsilon }(Ee^{sZ})^m)$$ where $Z=X_1Y_1$ is distributed according to a Normal Product distribution. I haven't carried out the calculation in full but I believe the moment generating function $Ee^{sZ}$ can be computed in close form using the expression (6) for $K_0$ found here.