So $\zeta=-\xi$, where $\xi$ is a primitive $p$'th root of unity. Let $\pi=1-\xi$. The minimal polynomial of $(1+\zeta^n)a$ is a product of terms of the form $X-(1+\zeta^{nj})\alpha_j$, where $j$ is odd, since the Galois conjugates of $\zeta$ are the odd powers of $\zeta$. Reducing mod $\pi$, you get a product of terms of the form $X-(1+(-\xi)^{nj})\alpha_j$. But $$ X-(1+(-\xi)^{nj}) \equiv X-(1+(-1)^{nj})\alpha_j \pmod\pi. $$ So if $n$ is odd, the minimal polynomial of $(1+\zeta^n)a$, reduced mod $\pi$, is a power of $X$, and since the polynomial has $p$-integral rational coefficients, they are all divisible by $p$. (Hmmm... I guess I've assumed that $a$ is $p$-integral. If $a$ has negative $p$-adic valuation, a similar argument should work, where now one may find that the constant term of the minimal poly in $\mathbb Z[X]$ is the one not divisible by $p$.)

So $\zeta=-\xi$, where $\xi$ is a primitive $p$'th root of unity. Let $\pi=1-\xi$. The minimal polynomial of $(1+\zeta^n)a$ is a product of terms of the form $X-(1+\zeta^{nj})\alpha_{ij}$, where $j$ is odd and $i$ runs over some index set (maybe depending on $j$). Note that $j$ is odd because the Galois conjugates of $\zeta$ are the odd powers of $\zeta$. Reducing mod $\pi$, you get a product of terms of the form $$ X-(1+(-\xi)^{nj}) \equiv X-(1+(-1)^{nj})\alpha_{ij} \equiv X-(1+(-1)^{n})\alpha_{ij}\pmod\pi. $$ So if $n$ is odd, the minimal polynomial of $(1+\zeta^n)a$, reduced mod $\pi$, is a power of $X$, and since the polynomial has $p$-integral rational coefficients, they are all divisible by $p$. (Hmmm... I guess I've assumed that $a$ is $p$-integral. If $a$ has negative $p$-adic valuation, a similar argument should work, where now one may find that the constant term of the minimal poly in $\mathbb Z[X]$ is the one not divisible by $p$.)