= Solution
Use the current convention in the question, whose vertex prefactor is $g/(2\cos\theta_W)$. The <neutral-current vector and axial couplings> are then
$$
c_V=T_3-2Q\sin^2\theta_W,\qquad c_A=T_3.
$$
For a charged <lepton>, $T_3=-1/2$, $Q=-1$, giving
$$
\boxed{c_V=2\sin^2\theta_W-\tfrac12,\qquad c_A=-\tfrac12.}
$$
For an active <neutrino>, $T_3=1/2$, $Q=0$, so $c_V=c_A=1/2$. The right-handed weak singlet contribution is purely vectorlike in the charge term; combining it with the left-handed contribution gives precisely this Dirac-current form. These coefficients would be halved if the vertex prefactor were $g/\cos\theta_W$ instead.
Let $p=k+k'$ be the <Z boson> momentum and $k,k'$ the outgoing massless <lepton> momenta. The <decay amplitude> is
$$
\mathcal M=\frac{g}{2\cos\theta_W}\epsilon_\mu(p)\bar u(k)\gamma^\mu(c_V-c_A\gamma_5)v(k').
$$
Use the <fermion spin sums> $\sum u\bar u=\not k$ and $\sum v\bar v=\not k'$. Anticommutation of $\gamma_5$ with the <Dirac gamma matrices> gives the symmetric part of the trace,
$$
L_{\rm sym}^{\mu\nu}=4(c_V^2+c_A^2)[k^\mu k'^\nu+k^\nu k'^\mu-g^{\mu\nu}k\cdot k'].
$$
The vector-axial cross term is antisymmetric in $\mu,\nu$, and hence drops out of the <massive vector polarization sum>. The massless equations also give $p_\mu L^{\mu\nu}=0$: in the symmetric part, the first two contractions sum to $p^\nu k\cdot k'$, canceling the last; the antisymmetric term vanishes because $p=k+k'$. The longitudinal $p_\mu p_\nu/m_Z^2$ term therefore makes no contribution.
Since $2k\cdot k'=m_Z^2$, contraction with $-g_{\mu\nu}$ gives $4(c_V^2+c_A^2)m_Z^2$. Thus the <massless fermion Z decay spin sum>, including all three initial polarizations, is
$$
\boxed{\sum_{Z,\ell,\bar\ell\ {
m spins}}|\mathcal M|^2=\frac{g^2m_Z^2}{\cos^2\theta_W}(c_V^2+c_A^2).}
$$
To use this fully summed expression for the decay width of one unpolarized <Z boson>, apply the <massive-vector spin average>, dividing by three. The width of each fixed polarization is the same after integration over all directions, so that procedure also gives the physical single-particle width. Omitting the division would sum the widths of three initial states.
For two massless distinguishable final particles, the <two-body Lorentz-invariant phase space> is
$$
d\Phi_2=\frac{d\Omega}{32\pi^2},\qquad \int d\Phi_2=\frac1{8\pi}.
$$
Consequently
$$
\Gamma=\frac1{2m_Z}\frac13\sum|\mathcal M|^2\frac1{8\pi}=\frac{g^2m_Z}{48\pi\cos^2\theta_W}(c_V^2+c_A^2).
$$
The tree-level <electroweak symmetry breaking> relation is $m_W=m_Z\cos\theta_W$. Combining it with the <Fermi constant> normalization gives
$$
\boxed{\Gamma_{Z\to\ell\bar\ell}=\frac{G_Fm_Z^3}{6\sqrt2\pi}(c_V^2+c_A^2).}
$$
For an active <neutrino>, the vertex itself projects out inactive chiralities, so no additional helicity factor is needed. This is the tree-level <massless leptonic Z decay width>; lepton masses and radiative corrections have been excluded.
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