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Let $d = \phi(n)$ and assume $n>2$.Then the primitive $n$-th roots of unity occur in $d/2$ complex conjugate pairs, and the GM-AM inequality applied to the (positive) contributions from each pair gives $\Phi_n(p)/p^d \leq (1 + \frac{1}{p^2}+ \frac{2}{dp})^{d/2}$, since the sum of the primitive $n$-th roots of unity is at most $1$. This is less than $e^{1/p}(1+ \frac{1}{p^{2}})^{d/2}$.Second term in product is at most $ e^{d/2p^{2}}$,so the quotient you are interested in is at most $e^{1/p + d/2p^{2}}$, as compared with $\sum_{j=0}^{\infty} 1/p^{j}$. This only helps when $p$ is fairly large compared to $n$ though.

Let $d = \phi(n)$ and assume $n>2$.Then the primitive $n$-th roots of unity occur in $d/2$ complex conjugate pairs, and the GM-AM inequality applied to the (positive) contributions from each pair gives $\Phi_n(p)/p^d \leq (1 + \frac{1}{p^2}+ \frac{2}{dp})^{d/2}$, since the sum of the primitive $n$-th roots of unity has absolute value at most $1$. This is less than $e^{1/p}(1+ \frac{1}{p^{2}})^{d/2}$.Second term in product is at most $ e^{d/2p^{2}}$,so the quotient you are interested in is at most $e^{1/p + d/2p^{2}}$, as compared with $\sum_{j=0}^{\infty} 1/p^{j}$. This only helps when $p$ is fairly large compared to $n$ though.

Let $d = \phi(n)$ and assume $n>2$.Then the primitive $n$-th roots of unity occur in $d/2$ complex conjugate pairs, and the GM-AM inequality applied to the (positive) contributions from each pair gives $\Phi_n(p)/p^d \leq (1 + \frac{1}{p^2}+ \frac{2}{dp})^{d/2}$, since the sum of the primitive $n$-th roots of unity is at most $1$. This is less than $e^{1/p}(1+ \frac{1}{p^{2}})^{d/2}$.Second term in product is at most $ e^{d/2p^{2}}$,so the quotient you are interested in is at most $e^{1/p + d/2p^{2}}$, as compared with $\sum_{j=0}^{\infty} 1/p^{j}$.

Let $d = \phi(n)$ and assume $n>2$.Then the primitive $n$-th roots of unity occur in $d/2$ complex conjugate pairs, and the GM-AM inequality applied to the (positive) contributions from each pair gives $\Phi_n(p)/p^d \leq (1 + \frac{1}{p^2}+ \frac{2}{dp})^{d/2}$, since the sum of the primitive $n$-th roots of unity is at most $1$. This is less than $e^{1/p}(1+ \frac{1}{p^{2}})^{d/2}$.Second term in product is at most $ e^{d/2p^{2}}$,so the quotient you are interested in is at most $e^{1/p + d/2p^{2}}$, as compared with $\sum_{j=0}^{\infty} 1/p^{j}$. This only helps when $p$ is fairly large compared to $n$ though.

This should help, but it's late here:let $d = \phi(n)$ and assume $n>2$.Then the primitive $n$-th roots of unity occur in $d/2$ complex conjugate pairs, and the GM-AM inequality applied to the (positive) contributions from each pair gives $\Phi_n(p)/p^d \leq (1 + \frac{1}{p^2}+ \frac{2}{dp})^{d/2}$, since the sum of the primitive $n$-th roots of unity is at most $1$. This is less than $e^{1/p}(1+ \frac{1}{p^{2}})^{d/2}$.Second term in product is at most $ e^{d/2p^{2}}$,so the quotient you are interested in is at most $e^{1/p + 2/p^{2}}$, as compared with $\sum_{j=0}^{\infty} 1/p^{j}$.

Let $d = \phi(n)$ and assume $n>2$.Then the primitive $n$-th roots of unity occur in $d/2$ complex conjugate pairs, and the GM-AM inequality applied to the (positive) contributions from each pair gives $\Phi_n(p)/p^d \leq (1 + \frac{1}{p^2}+ \frac{2}{dp})^{d/2}$, since the sum of the primitive $n$-th roots of unity is at most $1$. This is less than $e^{1/p}(1+ \frac{1}{p^{2}})^{d/2}$.Second term in product is at most $ e^{d/2p^{2}}$,so the quotient you are interested in is at most $e^{1/p + d/2p^{2}}$, as compared with $\sum_{j=0}^{\infty} 1/p^{j}$.