Let $k$ be a local field of characteristic $p$ and $\omega \in k$ a uniformiser. Consider the Artin-Schreier extension $L_n = k[x]/(x^p - x - \omega^n)$ for each $n \in \mathbb{Z}$.
Is there an explicit description for the image of the norm map $N_{L_n/k}: L_n^\times \to k^\times$?
If $n < 0$ it seems that the extension is wildy ramified and I'm not sure how to calculate the image of the norm group in this case.
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 $b \in k^\times$.
Well, I can give a very satisfying answer to a request you didn’t make, to find something whose norm is trivial, a super-partial answer to the question of identifying at least some non-norms, and a way of looking at the situation that may help you.
First, I’m going to call the chosen uniformizer $t$ rather than $\varpi$ or $\pi$ or $\omega$: too much writing. I’ll be thinking of $k$ as $\kappa((t))$ for a constant ring $\kappa=\Bbb F_q$ for $q=p^s$. At the end, I’ll specialize to the case $\kappa=\Bbb F_p$. We can always let $t$ be the chosen uniformizer in the ring of local integers $\kappa[[t]]$. And I’ll make $n>0$, so that our polynomial is $X^p-X-t^{-n}$. We need $(p,n)=1$ to guarantee irreducibility of this polynomial.
Normalizing the valuation of $t$ to $1$, we get the valuation of a root $1/\lambda$ of the polynomial to be $-n/p$. Then $\lambda$ won’t be a uniformizer except in the case $n=1$, but its minimal polynomial will be $$X^p+t^nX^{p-1}-t^n\,,$$ and the minimal polynomial of $1+\lambda$ will be $$ (X-1)^p+t^n(X-1)^{p-1}-t^n=X^p+t^n(X^{p-1}+X^{p-2}+\cdots+X)-1\,. $$ Thus $\mathbf N^{L_n}_k(1+\lambda)=1$.
Back in the case $n=1$, $\lambda$ is a uniformizer, and you can convince yourself that $\mathbf N^{L_1}_k(1+(\lambda^2))\subset1+(t^2)$, so that no norm starts out $1+t+\cdots$ . Everything else in the group of principal units $1+(t)$ should be a unit, however, in particular $(1+t)^p=1+t^p$. As you see, I’ve left a lot to be verified, but I believe there will be no problems.
But to understand what’s happening, you must have as accurate as possible a grasp of the structure of the multiplicative group $1+(t)$. First, it’s a $\Bbb Z_p$-module, because any series $1+$ (higher terms in $t$)may be raised to a power that’s a $p$-adic integer. Take your $p$-adic integer $z$, and express it in the form $z=\sum_n a_np^n$, where the $a_n$ are natural numbers. Then in $1+(t)$, the powers $(1+t)^{p^n}$ approach $1$, and $\prod_n\bigl((1+t)^{p^n}\bigr)^{a_n}$ is a convergent product.
Not only that, but you see that the binomials $1+t^m$ for $\gcd(p,m)=1$ are independent in the whole module $1+(t)$, and in the very special case that the constant field $\kappa$ is $\Bbb F_p$, that every element $g\in1+(t)$ may be written $$ g=\prod_{\gcd(p,m)=1}(1+t^m)^{z_m}\,, $$ and uniquely so. I won’t go into the general case here, but the dreaded Artin-Hasse Exponential can give some help. It’s all in Hazewinkel’s Formal Groups and Applications, anyway.
Back in the special case $\kappa=\Bbb F_p$, one may get a little insight by writing out the norms of all the $1+t^m$ for $(p,m)=1$ and seeing what’s missing. This is surely not the right approach, however.
