Local class field theory has something to say about the norm groups. To set up notation, let's put $k = \kappa((t))$ with uniformiser $\omega = t$. I'll assume $n < 0$ is not divisible by $p$ in the following (without loss).
The element $x \in L_n$ satisfying $x^p - x = t^n$ has valuation $v(x) = n/p$ (where I always normalise the valuation to satisfy $v(t) = 1$). For integers $a, b$ satisfying $ap + bn = 1$, the element $\pi := t^a x^b$ is now a uniformiser of $L_n$. If $\sigma$ is the generator of $G := \operatorname{Gal}(L_n/k)$ sending $x$ to $x+1$, a calculation shows that $v(\sigma(\pi) - \pi) = (1-n)/p$. Hence for the ramification groups in lower numbering we have $G_{-n} =G$, $G_{-n+1} = 1$. For the upper numbering we still have $G^{-n} = G$, $G^{-n+1} = 1$ (the renumbering is trivial up to the relevant point). (See for instance Section II.10 of Neukirch's Algebraische Zahlentheorie for the definition of these groups.)
It follows that the ideal $(t)^{1-n}$ is precisely the conductor of the extension $L_n/k$, and so the subgroup $1 + (t)^{1-n} \leq k^\times$ is contained in $N_{L_n/k}(L_n^\times)$, but $1 + (t)^{-n}$ is not. (See again Neukirch, Sections 1 and 6 of Chapter V.) If $\kappa = \mathbb{F}_p$ is the prime field, then $1 + (t)^{1-n}$ has index $p$ in $1 + (t)^{-n}$, showing that the only elements of $1 + (t)^{-n}$ which are norms from $L_n$ are the ones in $1 + (t)^{1-n}$. Note that this matches Lubin's answer for $n=-1$, and satisfies the desideratum (from the comments to the question) of finding many non-norms in $k$.
I do not know if one can easily identify the norm group more precisely than this. Serre's Corps Locaux has some more material, starting in V.3. The upshot there seems to be that one has a good handle on the norm in the graded components of $\kappa[[t]]^\times$, where $\kappa[[t]]^\times$ is filtered by the subgroups $1 + (t)^m$. This gives little more information than the above, however (it is precisely the above for $\kappa = \mathbb{F}_p$). Piecing things together for an ungraded version is presumably going to be tedious.