More generally, allow $p$ to be any prime and let $K$ be a finite extension of $\mathbf{Q}_p$ containing a primitive $p$-th root $\zeta$ of $1$ (which is automatic when $p=2$). The ramification index $e$ of $K$ over $\mathbf{Q}_p$ is divisible by $p-1$ (because $K$ contains $\mathbf{Q}_p(\zeta)$ by hypothesis); define $e_1$ by $e=(p-1)e_1$.

A little work as in Chapter 15 of Hasse's *Number Theory* (or as in Section V of Local discriminants) allows you to determine the structure of the filtered $\mathbf{F}_p$-space $K^\times\!/K^{\times p}$. The filtration on this quotient comes from the filtration
$$
\cdots U_2\subset U_1\subset\mathfrak{o}_K^\times\subset K^\times
$$
on $K^\times$, where $\mathfrak{o}_K$ is the ring of integers of $K$, with unique maximal ideal $\mathfrak{p}_K$, and, for every $i>0$, $U_i=1+\mathfrak{p}_K^i$ is the kernel of $\mathfrak{o}_K^\times\to(\mathfrak{o}/\mathfrak{p}_K^i)^\times$. Denote the image of $U_i$ in $\overline{K^\times}=K^\times\!/K^{\times p}$ by $\bar U_i$. Then the image of $\mathfrak{o}_K^\times$ is $\bar U_1$,
we have $\bar U_{pe_1+1}=\{1\}$, and the filtration on $\overline{K^\times}$ looks like
$$
\{1\}
\subset_1\bar U_{pe_1}
\subset_f\bar U_{pe_1-1}
\cdots
\subset_f\bar U_{pi+1}
=\bar U_{pi}
\subset_f\cdots
\subset_f\bar U_1
\subset_1\overline{K^\times}.
$$
Here, $i$ is any integer in the interval $[1,e_1[$ (which is empty when
$e_1=1$), an inclusion $E\subset_rE'$ means that $E$ is a codimension-$r$ subspace of $E'$, and $f$ is the residual degree of $K$ over $\mathbf{Q}_p$.

We have the hilbertian pairing $\overline{K^\times}\times\overline{K^\times}\to{}_pK^\times$, where ${}_pK^\times$ is of course the group of $p$-th roots of $1$ in $K$.

*The orthogonal complement of the subspace $\bar U_i$ for the hilbertian pairing is precisely $\bar U_{pe_1-i+1}$, for every $i\in[0,pe_1+1]$, provided we adopt the convention $\bar U_0=\overline{K^\times}$.*

It is amusing to try to figure out the analogue of all this when $K$ is a finite extension of $\mathbf{F}_p((\pi))$, where $\pi$ is transcendental.

**Addendum 1** Okay, here is a brief sketch of the proof. First, before the hilbertian pairing, there is the kummerian pairing :
$$
\overline{K^\times}\times\mathrm{Gal}(M|K)\to{}_pK^\times,
$$
where $M$ is the maximal abelian extension of $K$ of exponent $p$. The group $G=\mathrm{Gal}(M|K)$ comes with a natural filtration : the ramification filtration in the upper numbering. One may ask : how is the filtration on $\overline{K^\times}$ related to the filtration on $G$ ? *Answer :* The two filtrations are orthogonal to each other in an appropriate sense. See for example Section IX of Local discriminants.

Secondly, we have the reciprocity isomorphism $\rho:\overline{K^\times}\to G$ (with a normalisation which doesn't affect anything here), and the hilbertian pairing is obtained from the kummerian pairing via this isomorphism. Moreover, the filtration on $G$ is the image of the filtration on $\overline{K^\times}$ by $\rho$. Putting these two things together gives you the result.

**Addendum 2** What happens when the local field $K$ has characteristic $p$ ? Kummer theory has to be replaced by Artin-Schreier theory, so we have to first understand the filtration on $\overline{K^+}=K^+/\wp(K^+)$, where $K^+$ is the additive group of $K$ and $\wp(x)=x^p-x$. Denoting the image of $\mathfrak{p}_K^i$ by $\overline{\mathfrak{p}^i}$, it turns out that $\overline{\mathfrak{p}}=\{0\}$, and the analogous picture is
$$
\{\bar0\}\subset_1
\overline{\mathfrak{p}^0}\subset_f
\overline{\mathfrak{p}^{-1}}
\cdots\subset_f
\overline{\mathfrak{p}^{pj+1}}
=
\overline{\mathfrak{p}^{pj}}
\subset_f
\overline{\mathfrak{p}^{pj-1}}
\cdots\subset K^+\!/\wp(K^+).
$$
See for example Further remarks.

Let $M$ be the maximal abelian extension of $K$ [**edit** of exponent $p$] and $G=\mathrm{Gal}(M|K)$. We have the analogous pairing
$$
\overline{K^+}\times G\to\mathbf{F}_p,
$$
and we still have the ramfication filtration (in the upper numbering) on $G$. It turns out that the two filtrations are orthogonal to each other under this pairing.

As before, putting $\overline{K^\times}=K^\times\!/K^{\times p}$, we have the reciprocity isomorphism $\rho:\overline{K^\times}\to G$, and it carries the filtration on $\overline{K^\times}$ (which is no longer finite) onto the filtration on $G$.

Putting these two facts together gives the analogous result in characteristic $p$. I leave for you the pleasure of working out the details.

**Addendum 3** (2016/09/06) Still not convinced ? Some more details can be found in my Note arXiv:1609.01160.