TL;DR: OP's reference to Noether's theorem is exactly right. Angular momentum $x^{[i}p^{j]}$ is the Noether charge for rotational symmetry.
The Noether charge in point mechanics reads $$ Q~=~ p_i X^i - h T, \tag{1}$$ where $X^i$ and $T$ are spatial and temporal generators, and where $p_i=\frac{\partial L}{\partial v^i}$ is the momentum and $h=p_iv^i-L$ is the energy.
The $n$-dimensional rotation group $SO(n)$ has a Lie algebra $$so(n)~=~\{\omega\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \omega^T=-\omega\} \tag{2}$$ consisting of antisymmetric matrices. It generates infinitesimal rotations.
An infinitesimal spatial rotation $$ \delta x^i ~=~\epsilon X^i, \qquad \delta t~=~\epsilon T, \tag{3}$$ (where $\epsilon$ is an infinitesimal parameter), has generatorgenerators$^1$ $$ X^i = \omega^i{}_jx^j, \qquad T~=~0,\tag{4}$$ so$$ X^i = \omega^i{}_jx^j, \qquad T~=~0.\tag{4}$$
So the corresponding Noether charge is $$ Q~\stackrel{(1)+(4)}{=}~p_i \omega^i{}_jx^j~=~\underbrace{x^{[j}p^{i]}}_{\text{ang. mom.}}\omega_{ij}.\tag{5} $$
See also e.g. this related Phys.SE post.
--
$^1$ We raise and lower spatial indices with the spatial metric $\delta_{ij}$.