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.

Edit: In the comments, Daniel very reasonably asks to characterise the elements of $1 + (t)^{-n}$ which do occur as norms, in the case of a general finite residue field $\kappa$. Here the graded idea does help (see Proposition 5(iii) in V.3 in Serre): Restricting to the graded component $(1 + (t)^{-n})/(1 + (t)^{-n+1})$ of $k^\times$ and identifying it with the additive group $\kappa$ (sending $c \in \kappa$ to the class of $1 + ct^{-n}$), the image of the norm map is the image of an additive polynomial $\alpha X^p + \beta X \in \kappa[X]$. However, we know that this polynomial must in fact be defined over $\mathbb{F}_p$ (since everything we do arises via base change from $\mathbb{F}_p$, see also Serre's discussion in V.4) and not injective on $\mathbb{F}_p$. Hence it is a constant multiple of the Artin-Schreier polynomial $X^p - X$, and can be written as $(\beta X)^p - \beta X$ with $\beta \in \mathbb{F}_p^\times$. This confirms Daniel's suspicion that an element $1 + ct^{-n} + ... \in k^\times$ is a norm if and only if $c$ has the form $d^p - d$ for some $d \in \kappa$.

Edit, later: I somehow missed that Serre actually gives an explicit description of the norms later, in XIV.5. Specifically, he defines a local symbol $[a, b)_v \in \mathbb{Z}/p$ for $a \in k$, $b \in k^\times$, which is essentially equivalent to computing the local Artin symbol in the cyclic Artin-Schreier extension corresponding to $a$.
Relevant for us is the fact that $[t^n, b)_v = 0$ iff $b$ is a norm of the extension $L_n/k$. Further, Serre gives the explicit formula (Corollaire to Proposition 15) $[a, b)_v = \operatorname{Tr}_{\kappa/\mathbb{F}_p}(\operatorname{Res}(a \cdot \mathrm{d}b/b))$, where the residue of a differential is defined in the usual way (write the differential as $f \mathrm{d}t$ for some Laurent series $f$, and extract its $t^{-1}$-coefficient.)
From this formula we see that an element $b = 1 + ct^n + \dots$ is a norm in $L_n/k$ iff $\operatorname{Tr}_{\kappa/\mathbb{F}_p}(c) = 0$; the latter condition is equivalent to $c$ having the form $d^p - d$ by additive Hilbert 90.
The advantage of the formula over what's written above is of course that it also allows to treat a general element $b \in k^\times$.