Status of PP-completeness of MAJ3SAT
SHORT ANSWER:
It is unknown whether $\mathsf{MAJ3CNF}$ is a $\mathsf{PP}$-complete problem under many-one reductions. 



LONG ANSWER:
First of all, you refer to Bailey, Dalmau and Kolaitis, and their work on "Phase Transitions of $\mathsf{PP}$-complete Satisfiability Problems" in your question. Let me quote them:

'It is also worth noting that, although $\mathsf{MAJORITY \ SAT}$ is  $\mathsf{PP}$-complete, it is not known whether there is an integer $k \geq 3$, such that $\mathsf{MAJORITY}$ $\mathsf{kSAT}$ is $\mathsf{PP}$-complete.'

[http://www.sciencedirect.com/science/article/pii/S0166218X06004665 ]

It is, indeed, correct that Cook-Levin reduction is parsimonious and produces a 3CNF from a given CNF. As $\mathsf{\#CNF}$ is $\mathsf{\#P}$-complete, it immediately follows that $\mathsf{\#3CNF}$ is also $\mathsf{\#P}$-complete under parsimonious reductions. However, as already pointed in a comment, parsimonious reductions do not preserve majority. These reductions introduce auxiliary variables to reduce the size of the clauses, but in turn, these auxiliary variables increase the total number of assignments. For instance, consider the 4CNF which consists of a single clause:

$
\phi = (x_1 \vee x_2 \vee x_3 \vee x_4)
$

which is transformed into

$
\phi'=(x_1 \vee x_2 \vee y) \wedge (y \leftrightarrow (x_3 \vee x_4)) 
$

using the auxiliary variable $y$ and finally to the 3CNF

$
\psi= (x_1 \vee x_2 \vee y) \wedge (\neg y \vee x_3 \vee x_4) \wedge (y \vee \neg x_3 ) \wedge (y \vee \neg x_4). 
$

This transformation clearly preserves the model count, but it is easy to see that the majority is not preserved. $\phi$ has 15 satisfying assignments out of 16 assignments whereas $\psi$ has 15 satisfying assignments out of 32 assignments. In the former, majority satisfiability holds whereas in the latter  majority satisfiability does not hold.

Therefore, NO, the proof that #3CNF is $\mathsf{\#P}$-complete can not be adapted to prove that $\mathsf{MAJ3CNF}$ is $\mathsf{PP}$-complete? It remains open whether $\mathsf{MAJ3CNF}$ is a $\mathsf{PP}$-complete problem under many-one reductions. 

$\mathsf{MAJ3CNF}$ does not give much of an insight between the differences of $\mathsf{\#P}$ and $\mathsf{PP}$. Actually decision decision variant of $\mathsf{\#3CNF}$, say $\mathsf{D\#3CNF}$, is defined as follows: given a CNF $\phi$ and a number $m \geq 0$, decide whether $\phi$ has at least $m$ satisfying assignments. Note that for $\mathsf{D\#3CNF}$, we do not care about majority. Thus, we can transform any CNF into 3CNF using a parsimonous reduction, which proves that $\mathsf{D\#3CNF}$ is $\mathsf{PP}$-complete under many-one reductions. $\mathsf{MAJ3CNF}$ is simply a different problem than $\mathsf{D\#3CNF}$.