$K^{*n}$ has finite index in $K^*$ for an infinite extension field $K$ of $\mathbb{Q}_p$?

For $K \subset \overline{\mathbb{Q}}_p$ the index of $K^{\times n}$ may or may not be infinite. Clearly for $K = \overline{\mathbb{Q}}_p$ everything is an $n$th power so $K^{\times n} = K^{\times}$. On the other hand, let $K = \mathbb{Q}^{\mathrm{unr}}_p$ be the maximal unramified extension of $\mathbb{Q}_p$ with ring of integers $A$, so $p$ is prime in $A$. Then $1+px$ and $1+py$ are distinct elements of $K^{\times}/K^{\times p}$ whenever $x$ is not congruent to $y \bmod p$, because any $p$th power in $K$ which is $1 \bmod p$ is also $1 \bmod p^2$. Since the residue field is infinite, so is $K^{\times}/K^{\times p}$.

You meant $K\subset A=\overline{\Bbb{Q}}_p$. Without this condition take $K =A(x)$ for any $x\not \in A$,

$K^{*}/A^{*}\cong \Bbb{Z}[A]$ (send $f(x)/g(x)$ to $\sum_{f(a)=0} [a]-\sum_{g(b)=0} [b]$)

and $K^{*n}/A^{*}\cong n\Bbb{Z}[A]$

so that $$K^{*}/ K^{*n}\cong \Bbb{Z}[A]/n\Bbb{Z}[A]$$

Also for $p\nmid n$ and $K=\bigcup_j K_j$ where $K_j$ is an increasing sequence of finite extensions, from the Binomial series $(1+\pi_{K_j}O_{K_j})^n=1+\pi_{K_j}O_{K_j}$, and since $$K_j^* = \pi_{K_j}^{\Bbb{Z}}\langle \zeta_{p^{f_j}-1} \rangle(1+\pi_{K_j}O_{K_j})$$ we get $$[K_j^*:K_j^{*n}]=n \gcd(n,p^{f_j}-1),\qquad [K^*:K^{*n}]\le n^2$$