Let me first make a comment on Jorgensen's work. His Annals paper that you refer to was based on computations he first made in the cusped case, but did not appear in print until recently. He then did "orbifold Dehn filling" to get a closed example, which he was then able to publish explicitly, without revealing the origin of his techniques. I believe Thurston was inspired by Jorgensen's examples to prove his geometrization theorem for fibered manifolds. One may see pictures of the Ford domains and limit sets for the cusped case using the program Opti. Unfortunately, though, you can't draw the types of "snowflake curves" with this program, unless maybe you know the coordinates of group precisely.

Given a mapping class of a punctured torus, you get a corresponding map of its representation variety. If you want the commutator to lie in a specific conjugacy class (like an elliptic of order 3), then you get a very specific equation for the traces of elements in the group related to the Markoff equation, which is given in Equation 3.3 in McMullen's book (for derivation of these types of relations in $SL(2,C)$, you could refer to Chapter 1 of Marden's book on Kleinian groups).

So now you get an action of the mapping class on the traces, which McMullen computes explicitly for the transformation $LR$, and will be the same algebraic transformation regardless of the value of $tr[A,B]$.

McMullen computes (see Marden) $$L(\alpha, \beta,\gamma)=(\gamma, \beta, \beta\gamma-\alpha),$$ $$R(\alpha,\beta,\gamma)=(\alpha,\gamma,\alpha\gamma-\beta).$$

These preserve the Markoff equation $\alpha^2+\beta^2+\gamma^2=\alpha\beta\gamma+3$ which holds in your case of an elliptic commutator of order 3. Solving for the fixed point $LR(\alpha,\beta,\gamma)=(\alpha\gamma-\beta,\gamma,\gamma^2\alpha-\beta\gamma-\alpha)=(\alpha,\beta,\gamma)$, we obtain the fixed point $(\alpha,\alpha/(\alpha-1),\alpha/(\alpha-1))$, subject to the Markov equation. Plugging in, we see that $\alpha$ is a root of the equation $\alpha^4-3\alpha^3+6\alpha-3=0$.
This equation has two real roots, and two complex conjugate roots, either of which gives the desired fixed point (up to complex conjugation) giving the traces of a totally degenerate group.

Now, to compute the eigenvalue of the fixed point, we compute the characteristic polynomial of the derivative of $LR$ (which is a 3x3 matrix). I won't go through the computation, but I get the polynomial $\lambda^3-(2\alpha^2/(\alpha-1))\lambda^2 +2\alpha^2/(\alpha-1)\lambda-1$ after substituting for the fixed point. Dividing by the obvious factor $\lambda-1$ (which I think comes from the fact that the mapping class group has a conserved quantity $tr[A,B]$ on the character variety), we have your desired eigenvalue is a root of the polynomial $\lambda^2 - (2\alpha^2/(\alpha-1) +1)\lambda +1$.