Let $K$ be a local field of characteristics $0$. It is known that in that case $(K^*)^n$ is an open subgroup but is it of finite index? I suppose that in general case the answer is no, but what about finite extensions of $\mathbb Q_p$?
And the second question is: from local class field theory we know that if $x$ is the norm of every cyclic extension whose degree divides $n$ then $x$ is an $n$-th power but is it true in opposite direction?
On
Let $U$ be the units of $K^*$ (elements of valuation zero). Then $|K^*:K^{*n}|=n|U:U^n|$ so we need to show that $|U:U^n|<\infty$. By Hensel, if $a\equiv1\pmod{\pi_K^m}$ for some sufficiently large $m$, then $a$ is an $n$-th power. So $U^n\supseteq U_{(m)}$, the set of such $a$, which has finite index in $U$.
You don't give the precise definition of a local field $K$ (of characteristic $0$), but since you allude later to finite extensions of $\mathbf Q_p$ as a special case, I'll assume that your $K$ is generally a complete field w.r.t. a discrete valuation, with residual field $k$ of characteristic $p\neq 0$ (the so called "unequal characterictic" case). Then :
1) To study the possible finiteness of $K^*/{K^*}^n$, one first looks at the effect of taking $n$-th powers on the descending chain of subgroups $K^*> U_0=U >...> U_i >...$ , where $U$ is the group of units and $U_i =1+ (\pi)^i$, $\pi$ being an uniformizer. Concerning the successive quotients $U_i/U_{i+1}$, obviously $K^*/U \cong (\mathbf Z/n\mathbf Z, +)$ via valuation, and it is classically known that $U/U_1\cong (k^*, \times)$ and $U_i/U_{i+1} \cong (k,+)$ for $i \ge 1$ (see e.g. Cassels-Frôhlich, chap.I, prop.4); it follows at once from the first property that $K^*/{K^*}^n$ is finite iff $U/U^n$ is finite. Applying the snake lemma (relative to $n$-th powers) to the exact sequence $1 \to U_1\to U\to k^*\to 1$, one gets an exact sequence ... $\mu_n (k)\to U_1/({U_1)}^n \to U/U^n \to {k^*/k^*}^n \to 1$, where $\mu_n (k)$ is the (finite) group of $n$-th roots of $1$ in $k^*$. This shows that $U/U^n$ is finite iff $U_1/({U_1)}^n$ and ${k^*/k^*}^n$ are finite. The finiteness of ${k^*/k^*}^n$ condition, say @, already puts a restriction on the pair $(k, n)$. It is clear that @ is met for all $n$ if $k$ is finite; for all powers of $p$ if $k$ is perfect.
The 3rd property above can be exploited to control the effect of the $n$-th power map on the filtration of the $U_i/U_{i+1}$'s for $i\ge 1$:
(a) if $p$ does not divide $n$, the $n$-th power map is an automorphism of $U_i$ for all $i\ge 1$
(b) the $p$-th power map induces an isomorphism $U_i \cong U_{i+e}$ for $i>\frac e {p-1}$ where $e$ is the absolute ramification index of $K$ (see e.g. Serre's "Local Fields", chap.XIV, §4, prop.9)
These show that $K^*/{K^*}^n$ is finite for all $n$ if $k$ is finite; for all powers of $p$ if $k$ is perfect.
2) In the usual conditions of validity of local CFT (i.e. when $k$ is finite, which amounts to saying that $K$ is locally compact), the norm subgroups $M$ of $K^*$ are characterized by the so called existence theorem of CFT : $M$ is a norm subgroup of $K^*$ iff (i) the index $(K^*:M)$ is finite, and (ii) $M$ is open in $K^*$ (see Cassels-Fröhlich, chap.VI, §2.7, thm.3). Note that, if (i) is satisfied, (ii) is equivalent to "$M$ is closed". The proof of necessity is not difficult. The hard part is the proof of sufficiency. So ${K^*}^n$ is a norm subgroup. Its index can even be calculated explicitly using the machinery of Herbrand quotients, see e.g. Serre, op. cit., exercise 3).
In the particular setting of your second question, I suspect that perhaps you forgot the extra kummerian hypothesis, that $K$ contains the group $\mu_n$ of $n$-th roots of $1$, which allows to use the more precise apparatus of local Hilbert symbols (Serre, op. cit., prop.7). In particular, for $\bar a, \bar b \in K^*/{K^*}^n$, the symbol $<a,b>$ is $1$ iff $a$ is a norm in the cyclic extension $K(\sqrt[n] {b})/K$, and $<a,b>$ is $1$ for all $b$ iff $a \in {K^*}^n$ because the Hilbert symbol is non degenerate.