Here is a way to bound $n,m,l$ from which a computer can find the rest of the solutions then.
As Jack D'Aurizio points out it is equivalent to showing $\cos \frac{2\pi}{n} + \cos \frac{2\pi}{n} - \cos \frac{2\pi}{l} = 1$. Let $L = lmn$ and denote $\zeta_k = e^{2\pi i/k}$.
Consider the field $\mathbb{Q}[\zeta_L]$. Let $S$ denote the set of residues modulo $L$ that are relatively prime with $L$ and then denote $f_k$ the automorphism that sends $\zeta_L$ to $\zeta_L^k$ (if it exists. Then: $\sum_{k \in S} f_k(LHS) = \sum_{k \in S} f_k(RHS)$ Now, by using $2 \cos \frac{2\pi}{n} = \zeta_L^{ml} + \zeta_L^{-ml}$ one can show $\sum_{k \in S} f_k(\cos \frac{2\pi}{n}) = \frac{\phi(lmn)}{\phi(n)}\mu(n)$
Thus we require that $\frac{\mu(n)}{\phi(n)} + \frac{\mu(m)}{\phi(m)} - \frac{\mu(l)}{\phi(l)} = 1$
Which bounds $m,n,l$ quite nicely and a computer can finish from here.
If $n=2$ then note that $\frac{\mu(m)}{\phi(m)} - \frac{\mu(l)}{\phi(l)} = 2$ which is easy to check has no solutions because note that $\phi(m) = \phi(l) = 1$ is forced. Similarly $m=2$ has no solutions. Now, if $l=2$ then use the original equation of $L_n^2 + L_m^2 = L_l^2$ to deduce that $\frac{1}{m} + \frac{1}{n} = 1$ from which it is easy to find all solutions.
Now take $l,m,n > 2$. Note that then $\frac{\mu(n)}{\phi(n)} + \frac{\mu(m)}{\phi(m)} \le 1$
Thus for a solution to occur we need either $\mu(l) = 0$ or $\mu(l) = -1$. The $\mu(l) = 0$ case its easy to find $(m,n) = (6,6)$ is the only solution so let's take $\mu(l) = -1$.
We can assume now that $\mu(n) = \mu(m) = 1$ because otherwise by very similar stategies to above we can find all solutions otherwise. But then note that we have $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 2$
where $a = \phi(n)/2$, etc. This only has the solution where $a,b,c$ are a permutation of $(1,2,2)$ so from here we have bounded all the variables and we are done.