Actually, this never happens unless the nilradical is $0$ (i.e., unless the ring is a domain). Let $A$ be a ring with only one associated prime. Then $A$ has only one minimal prime, so the nilradical $N$ of $A$ is prime. Now localize at $N$. We now have a $0$-dimensional Noetherian local ring $A_N$, so it is artinian. In particular, $A_N$ has finite length over itself, and the length of the maximal ideal $NA_N$ is strictly smaller than the length of $A_N$. But we are assuming $N$ is locally free, so $NA_N$ is free over $A_N$; the only way this can happen is if $NA_N=0$. But, as you observe, every element of $A\setminus N$ is a nonzerodivisor, so the canonical map $A\to A_N$ is injective. Thus $NA_N=0$ implies $N=0$.

Actually, this never happens unless the nilradical is $0$ (i.e., unless the ring is a domain). Let $A$ be a Noetherian ring with only one associated prime whose nilradical $N$ is locally free. Then $A$ has only one minimal prime, so $N$ is prime. Now localize at $N$. We now have a $0$-dimensional Noetherian local ring $A_N$, so it is artinian. In particular, $A_N$ has finite length over itself, and the length of the maximal ideal $NA_N$ is strictly smaller than the length of $A_N$. But we are assuming $N$ is locally free, so $NA_N$ is free over $A_N$; the only way this can happen is if $NA_N=0$. But, as you observe, every element of $A\setminus N$ is a nonzerodivisor, so the canonical map $A\to A_N$ is injective. Thus $NA_N=0$ implies $N=0$.