Dear Doron,
Yes, Valentin Féray is right.
The asymptotic distribution of the sum of heights (also called total path
length, or internal path length) is the same for all these trees, and more
generally for the class of simply generated trees (where you may put in
weights depending on the degrees), as long as a certain variance is finite.
This was proved, by Aldous The continuum random tree II and III (London Math
Soc Lecture Notes 167, 1991 and Ann. Probab. 21 (1993), 248-289, based on
his idea that the contour of any such random tree converges (after
rescaling) to a Brownian excursion.
The limit distribution is thus the same for all such random trees, and
equals the area under a Brownian excursion.
This distribution is studied in my paper [201] Brownian excursion area, Wright's constants in graph enumeration, and
other Brownian areas. Probability Surveys 3 (2007), 80-145.
In Section 14 you will find a (not very nice) formula for the distribution
function as an infinite sum, due to Darling (1983).
For some related results, see my paper
[146] The Wiener index of simply generated random trees. Random Structures
Algorithms 22 (2003), no. 4, 337-358.
And as Valentin wrote, there is more on simply generated random trees (but
not about this sum) in my paper
[264] Simply generated trees, conditioned Galton-Watson trees, random
allocations and condensation. Probability Surveys, 9 (2012), 103-252.
You are also right that the same distribution appears for the number of
inversions in a 132-avoiding permutation. That is included in my paper
[290] Patterns in random permutations avoiding the pattern 132,
Combin. Probab. Comput., to appear. Longer version: arXiv:1401.5679
I was not aware of your work on this. (But the problem was inspired by our
paper.) You might like some of my other moment calculations in this paper
(in particular the arXiv version), although they are not automatic.
(My papers are also on my web page.)
Finally, I agree with Valentin that in this case, the distribution, and moments, are nicer if one does not centralize: the random variable is positive (and that's enough to see that it is not normal), but subtracting the mean destroys that. It may be simplest to normalize by dividing by its expectation, making the expectation = 1.
Best,
Svante
