Averaging i.i.d. variables: Equal chance to be right and left of the mean?

Yes it is possible for $c$ to take any value strictly between $0$ and $1$. The point is that there exist mean-zero stable distributions which are not symmetric about $0$ (of course, such a stable distribution cannot be Gaussian, and so it must have infinite variance). You may look at the Wikipedia page to see how some of these stable distributions look.

Specifically, if $\alpha \in (1,2)$ and $\beta \in [-1,1]$, then it turns out that there exists a random variable $X$ whose characteristic function will look like $$\phi_X(t) = e^{-|t|^{\alpha}\big(1-i\beta \tan(\frac{\pi\alpha}{2})\text{sign}(t)\big).}$$ As it turns out, this distribution will have mean zero, and moreover (by varying $\alpha$ and $\beta$), $P(X<0)$ can be any predefined number $c\in(0,1)$. Furthermore, for iid copies one may check directly from the characteristic function that $n^{-1/\alpha}(X_1+...+X_n)$ has the same distribution as $X_1$. From this we can easily conclude that $$P(L_n<0) =P(n^{1-1/\alpha}L_n<0)= P(X<0)=c \in (0,1),$$ for all $n$, as desired. I do not know if $c=0$ or $c=1$ is a possible limit for nonzero random variables $X_i$, though it'd be interesting to find out.

Partial answer: if $X_i$'s are non-negative with infinite mean than $L_n \to \infty$ a.s. an $P(L_n \leq x) \to 0$ for every $x$.