As was noted in the comments by Yuval and Kevin, even if $X_1$ is bounded, the best upper bound on the probability in question is a negative power of $\ln n$. To get such a bound (and even an asymptotics), it is actually enough to assume that $E|X_1|^k<\infty$ for some $k>2$. Indeed, a theorem due to S. Nagaev states this:
Suppose that $X_1,X_2,\dots$ are zero-mean unit-variance iid random variables, with $S_n:=\sum_1^n X_i$. Let $Z\sim N(0,1)$. Take any real $k>2$. Then the condition $E|X_1|^k<\infty$ is sufficient for the asymptotic relation $P(S_n\ge z\sqrt n)\sim P(Z\ge z)$ (as $n\to\infty$) to hold in the zone $0\le z\le\sqrt{(\frac k2-1)\ln n}$ and necessary for this relation to hold in the zone $0\le z\le\sqrt{(k+1)\ln n}$.
So, assuming that indeed $E|X_1|^k<\infty$ for some $k>2$, and letting $z=t\sqrt{2\ln\ln n}$, we see that $$P\Big(\Big|\frac{S_n}{\sqrt{2n\ln\ln n}}\Big|> t\Big) \sim P(Z\ge z)\sim\frac1{z\sqrt{2\pi}}e^{-z^2/2} =\frac1{2t\sqrt{\pi\ln\ln n}}(\ln n)^{-t^2} $$ for each $t>0$ as $n\to\infty$.