An $L\neq0$ state would necessarily be degenerate, however, the ground state (as long as no spin effects and/or velocity-dependent potentials are present) is always non-degenerate. To see this, consider the Ansatz
$$
\Psi(\boldsymbol{r}_1,...,\boldsymbol{r}_N)=\chi(\boldsymbol{r}_1,...,\boldsymbol{r}_N)e^{i\Phi(\boldsymbol{r}_1,...,\boldsymbol{r}_N)} \ ,
$$
where $\chi(\boldsymbol{r}_1,...,\boldsymbol{r}_N)\geq0$ without loss of generality.
You can always minimize the energy
$$
{\cal{E}}=\frac{\langle\Psi|H|\Psi\rangle}{\langle\Psi|\Psi\rangle}\geq E_0
$$
($E_0$ being the exact ground state energy)
by setting $\Phi(\boldsymbol{r}_1,...,\boldsymbol{r}_N)=\text{const.}$, which is equivalent to $\Phi(\boldsymbol{r}_1,...,\boldsymbol{r}_N)=0$. To verify this, just substitute, and then integrate by parts:
$$
{\cal{E}}=\frac{\langle\chi|H|\chi\rangle+\Delta}{\langle\chi|\chi\rangle} \ ,
$$
where
$$
\Delta=\frac{\hbar^2}{2m}\sum_{k=1}^{N}\int\mathrm{d}^3r_1...\int\mathrm{d}^3r_N
\chi^2(\nabla_k\Phi)^2\geq0 \ .
$$
You can always go lower in the energy by setting $\nabla_k\Phi=0$, that is $\Phi=\text{const.}$.
Therefore, the ground state wave function is always $\Psi\geq0$, meaning that no degenerate, yet orthogonal wave functions can be found (two degenerate wave functions could always be chosen orthogonal, however, this is not possible if none of them has nodes).
We can conclude that the ground state is non-degenerate, so it cannot be an $L\neq0$ state.
(This is just a trivial extension of the results of
Am. J. Phys, 70, 808 (2002)
for $N$ spinless particles.)
A note on two-electron systems
Although the above results are valid for spinless particles, we can still use them to prove without any computation that the ground state total spin of a two electron system is necessarily $S=0$. For a two-electron system, the wave function has a simple product form:
$$
\Phi_{S,M_S}(\boldsymbol{r}_1,\boldsymbol{r}_2)=\Psi(\boldsymbol{r}_1,\boldsymbol{r}_2) \, \Xi_{S,M_S} \ ,
$$
where $\Xi_{S,M_S}$ is a spinor with $2^2=4$ components, antisymmetric under interchanging the particle labels for $S=0$ and symmetric for $S=1$. We need a symmetric $\Psi(\boldsymbol{r}_1,\boldsymbol{r}_2)$ for $S=0$ and an antisymmetric one for $S=1$ in order to fulfill the Pauli principle. Apart from fixing the permutational symmetry properties of $\Psi(\boldsymbol{r}_1,\boldsymbol{r}_2)$, the spin does not have any effect, so we can use the previous results, and conclude that the spatial wave function does not have any nodes. This excludes the possibility of a permutationally antisymmetric spatial part, so the ground state must have $S=0$.
The ground state of any two-electron system is thus a spin-singlet with the spatial part being totally symmetric under the relevant point group ${\cal{G}}$:
$^1S$ for helium-like ions (where ${\cal{G}}=O(3)$),
$^1\Sigma_\text{g}^+$ for the H$_2$ molecule (where ${\cal{G}}=D_{\infty\text{h}}$),
and so on.
