The nilradical of a commutative ring $R$, denoted $Nil(R)$, is the set of all nilpotent elements in $R$. It can be shown that $Nil(R)$ is an ideal of $R$. Every ideal is equal to the kernel of some ring homomorphism: just consider the projection $\pi : R \to R/I$.
I'm wondering: is there a ring homomorphism whose kernel is $Nil(R)$ that is more explicit than the projection map $\pi : R \to R/Nil(R)$?
You're not going to get anywhere near something as crisp as the quotient $S_n/A_n$, because that is a highly special case: it's always order $2$ (for $n >1$) and cam be defined from the parity of the structure of the elements themselves.
In contrast, $R/Nil(R)$ is going to be quite wild by comparison. The quotient doesn't have a fixed order or any particularly special property, other than the fact it is a reduced ring.
But maybe the following is along the lines of what you're looking for. Let $\mathcal C$ be a family of prime ideals such that $\bigcap \mathcal C=Nil(R)$. Of course, it could be all prime ideals, but sometimes fewer will do.
You always have a homomorphism $R\to \prod_{P\in\mathcal C}R/P$, whose kernel is $Nil(R)$. By the first isomorphism theorem, $R/Nil(R)$ is a subring of the thing on the right, which is a product of domains. You could additionally even embed those domains in their fraction fields, and have a map of $R/Nil(R)$ into a product of fields.
Under special conditions, $R$ may permit selection of a finite $\mathcal C$ with the properties above, getting you an embedding into a finite product of fields: that's quite special.
If you know that there are finitely many maximal ideals that intersect to $Nil(R)$, then the Chinese Remainder Theorem actually says this injection is an isomorphism. In that case $R$ would be called a semilocal ring.
Conclusion
So anyway, now I hope you can see the diversity of structure of $R/Nil(R)$. I don't know if it is "more explicit," but it does tell you more about the structure of the quotient.