The mean square distance of a random walk from the origin

Let us divide the (time) interval $[0,n]$ into $n/t$ subintervals of length $t$. Let us call the $k$th interval good, if, during that interval, the random walk spends time at least $t/5$ to the left of $x_k-\sqrt{t}$ and at least $t/5$ to the right of $x_k+\sqrt{t}$ (here, $x_k=X_{kn/t}$ is the position of the walk in the beginning of the $k$th time interval). Clearly, the events ($\{k\mbox{th interval is good}\}$, $k=1,\ldots,n/t$) are independent, and there is $\alpha>0$ such that the probability that an interval is good is at least $\alpha$, uniformly in $t$ (you can prove this formally by e.g. comparing the random walk to the Brownian motion).

Next, a good interval contributes an amount at least $\frac{t}{5}\times (\sqrt{t})^2 = \frac{t^2}{5}$ to the sum. Moreover, a Chernoff's bound for the Binomial distribution implies that, with probability at least $1-\exp(-c' \frac{n}{t})$, at least $\frac{\alpha}{2}\times\frac{n}{t}$ intervals will be good. So, at least with the above probability, the sum will be at least $\frac{\alpha}{2}\times\frac{n}{t}\times \frac{t^2}{5} = \frac{\alpha t}{10n}n^2$. Now, take e.g. $t=n/\log^2n$ (or $t=n/(C\log n)$ for large $C$) to get a statement you're aiming at.