How can Majorana neutrinos have hypercharge? Left-handed leptons transform under the full SM gauge group $\text{SU}(3)_C\times\text{SU}(2)_L\times\text{U}(1)_Y$ as $(\textbf{1}, \textbf{2},-1/2)$, i.e. $(\nu_L,e_L)$ is an $\text{SU}(2)_L$ doublet.
If neutrinos were Majorana particles, they would be indistinguishable from their antiparticle counterparts, which means $\nu_L = \bar{\nu}_L$, but I thought the quantum numbers of $(\bar{\nu}_L,\bar{e}_L)$ were $(\textbf{1},\textbf{2},+1/2)$, as weak hypercharge and isospin flip when applying a C inversion.
What am I misunderstanding?
 A: To spare your sanity and mine, I won't talk about Majorana fermions, but, instead, about Majorana mass terms which violate lepton number by 2, while preserving Lorentz invariance. W.l.o.g., consider just one generation of leptons.

*

*The short answer to your question is you are discussing such terms below the symmetry breaking scale, where weak isospin and
hypercharge are broken/hidden/mooted. There is an invariant, the Weinberg operator, below, whose value in the SM vacuum is precisely your contested mass term.

Given $Q=T_3+ Y_w/2$, recall right-chiral  neutrinos $\nu_R$ are singlets under SU(2)×U(1). Now, indeed, charge conjugation reverses the weak hypercharge and isospin (eigenvalue of $T_3$) of
$$
L=\begin{pmatrix}\nu_L \\e^-_L \end{pmatrix},
$$
whose hypercharge is -1 and the isospin e'ves 1/2 and -1/2, respectively. I'm using the traditional normalization for the hypercharge, twice yours.
Thus, $L^c$ is a  R neutrino over a R positron, with opposite fermion number, hypercharge and isospin eigenvalues; so, strictly speaking, you are right that, unlike the Majorana mass term $m\overline{\nu_R} \nu_R^c$ which is a SM gauge singlet, $m\overline{\nu_L} \nu_L^c$ is not! It has w-hypercharge 2 and w-isospin eigenvalue -1.
However, the unrenormalizable dimension-5 Weinberg operator
$$
{y\over M} \overline{L}\cdot \tilde H ~~\tilde H \cdot L^c 
$$
is a singlet, since each of its factors is a singlet. Recall
$$
H= \begin{pmatrix}\Phi^+  \\ \Phi^0 \end{pmatrix}~~~~\leadsto \\ \tilde H =
 \begin{pmatrix}\Phi^{0 *}  \\ -\Phi^- \end{pmatrix}, 
$$
where the conjugate Higgs has w-hypercharge -1 and w-isospin 1/2 for the neutral component, which picks up the v.e.v.
Consequently, e.g., the v.e.v.
$$
\overline{L} \cdot \langle \tilde H\rangle ={1\over \sqrt 2} \overline{\begin{pmatrix}\nu_L \\e^-_L \end{pmatrix} } \cdot \begin{pmatrix} v  \\  0\end{pmatrix} ={v\over \sqrt 2} \overline{ \nu_L }
$$
is a de-facto low-energy singlet, unlike the contrary appearances before, just like its R sibling!!
This magical transmutation from non-singlet to singlet is a feature of symmetry breaking: at low energies, by dint of the  lopsided vacuum,  there is no weak isospin or hypercharge for many purposes. It is as though the vacuum ate up the hypercharge 2 and isospin -1 of that term.
You might enjoy this review.
