[This paper][1] shows that $L(n) > .061867\sqrt{n}$ for infinitely many $n$.

As for somewhat elementary methods (in the sense of avoiding the Riemann zeta function) to show that $L(n)$ is "usually" of order $\sqrt{n}$, one can use the Lambert series
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)q^n}{1-q^n}} = \sum_{n=1}^{\infty}{q^{n^2}}.$$

As
$$\frac{q^n}{1+q^n} = \frac{q^n}{1-q^n} - 2\frac{q^{2n}}{1-q^{2n}},$$
we have
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{q^{-n}+1}} = \sum_{n=1}^{\infty}{q^{n^2}} - 2\sum_{n=1}^{\infty}{q^{2n^2}}$$

or equivalently, letting $q = e^{-\pi/x}$ and $\psi(x) = \sum_{n=1}^{\infty}{e^{-\pi xn^2}}$, where $x$ is large,
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \psi(1/x) - 2\psi(2/x)$$

Now $\psi(x)$ satisfies the functional equation
$$\frac{1+2\psi(x)}{1+2\psi(1/x)} = \frac{1}{\sqrt{x}},$$

and so we can rewrite this as
$$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \frac{1-\sqrt{2}}{2}\sqrt{x} + \frac{1}{2} + (\psi(x)-2\psi(x/2))\sqrt{x}.$$

For large $x$, the left-hand side "looks like" $L(x)$, whereas the right-hand side is dominated by the term $\frac{1-\sqrt{2}}{2}\sqrt{x}$. This also explains why $L(n)$ is predominantly negative, as $\frac{1-\sqrt{2}}{2}$ is negative.

  [1]: http://www.davidson.edu/math/mossinghoff/LiouvilleSums2_BFM.pdf