For Helium atom we know there are spin singlet and spin triplet state corresponding to $S=0$ and $S=1$. But what if the electrons are more than two and what does singlet mean for orbital degree of freedom? For many-electron system, does $L=0$ mean an orbital singlet state, or does $L=2,4\ldots$ also mean an orbital singlet? And what's the relation between orbital singlet and symmetry of the orbital part wave function?
The terms singlet, doublet, triplet, quartet are names given to part of the Term Symbol in $LS$ coupling. It specifies the value of $S$, but in an encoded form where the left superscript is the numerical value of $2S+1$: $^1S$ denotes $S=0$. It's entirely historical: the classification of levels into terms and their naming was an empirical product of spectroscopy which long pre-dated quantum mechanics. Very old textbooks such as White Atomic Spectra, discuss the pre-quantum spectroscopic evidence in great detail before giving the quantum interpretation.
So to answer your specific questions: yes you can have singlet whenever $S=0$ is a possible result of adding up all the spins. This requires an even number of spins. And it is nothing to do with the value of L, the orbital angular momentum.