Actually $\zeta_{K_n}(\sigma)$ is bounded for any fixed $\sigma > 1$.

Let $N = 2^n = [K_n : {\bf Q}]$. Then all the local factors of $\zeta(\sigma)$, other than the factor $(1-2^{-\sigma})^{-1}$ for the prime above $2$, are of the form $(1 - q^{-\sigma})^{-g}$, where $q$ is a prime power congruent to $\pm 1 \bmod 4N$, and $g \mid N$. Thus $$ \zeta_{K_n}(\sigma) < \frac1{1-2^{-\sigma}} \prod_{m=1}^\infty \frac1{(1 - (2mN)^{-\sigma})^N} $$ and the product approaches $1$ as $n \to \infty$ because its logarithm behaves like $\sum_{m=1}^\infty N/(2mN)^\sigma = \zeta(\sigma) / N^{\sigma-1} \to 0$.

Since each local factor exceeds $1$, it follows that in fact $\zeta_{K_n}(\sigma) \to (1-2^{-\sigma})^{-1}$ as $n \to \infty$.

Actually $\zeta_{K_n}(\sigma)$ is bounded for any fixed $\sigma > 1$.

Let $N = 2^n = [K_n : {\bf Q}]$. Then all the local factors of $\zeta(\sigma)$, other than the factor $(1-2^{-\sigma})^{-1}$ for the prime above $2$, are of the form $(1 - q^{-\sigma})^{-g}$, where $q$ is a prime power congruent to $\pm 1 \bmod 4N$, and $g \mid N$. Thus $$ \zeta_{K_n}(\sigma) < \frac1{1-2^{-\sigma}} \prod_{m=1}^\infty \frac1{(1 - (2mN)^{-\sigma})^N} $$ and the product approaches $1$ as $n \to \infty$ because its logarithm behaves like $\sum_{m=1}^\infty N/(2mN)^\sigma = \zeta(\sigma) \, / \, 2^\sigma N^{\sigma-1} \to 0$.

Since each local factor exceeds $1$, it follows that in fact $\zeta_{K_n}(\sigma) \to (1-2^{-\sigma})^{-1}$ as $n \to \infty$.