The computation of $H^1(\operatorname{Spec}(k), \mu_n)$ is not completely correct: in Serres book it is written the correct answer $H^1( \operatorname{Gal}(k^s|k),\mu_n(k^s))= k^{\times}/ \mathord{{k^{\times}}^n}.$
Another possible way to prove it is to consider the sequence
$$0\to \mu_n \to \mathbb Gm \to \mathbb Gm \to 0,$$
where the map between the multiplicative groups is the "power-to-the-$n$". The sequence is exact for the etalé topology (but not for the Zarisky topology), hence you get a long exact sequence for the cohomology. But $ H^1(\operatorname{Spec}(k), \mathbb{G}m)=0$ for Hilbert's theorem 90, hence you get the result.
About the second case, so $H^1(\operatorname{Spec}(k), \underline{\mathbb{Z}/\mathord{n \mathbb{Z}}})$, you cannot reduce to the previous case since $\underline{\mathbb{Z}/\mathord{n \mathbb{Z}}}\not{\cong}\mu_n$ even if $k$ contains all $n$-roots of unity. But you can reduce to compute Galois cohomology as you did, so you want to compute $H^1(\operatorname{Gal}(k^s/k), \mathbb{Z}/\mathord{n \mathbb{Z}})$, where now we have a constant group, since the points over $k^s$ of the constant sheaf is the constant group (and over every field). But now we have group cohomology, where the group acts trivially, hence the $H^1$ is just the hom's, so
$$H^1(\operatorname{Gal}(k^s/k), \mathbb{Z}/\mathord{n \mathbb{Z}})\cong \operatorname{Hom}(\operatorname{Gal}(k^s/k),\mathbb{Z}/\mathord{n \mathbb{Z}}).$$
This last group can be big in general, but it is small for example if $k$ is finite: it is isomorphic to $\mathbb{Z}/\mathord{n \mathbb{Z}}$.
In down to earth terms, the elements of $H^1(\operatorname{Spec}(k), \underline{\mathbb{Z}/\mathord{n \mathbb{Z}}})$ correspond essentially to Galois extensions of $k$ with cyclic Galois group of order dividing $n$.
The main Conjecture (MC) has at least 3 different equivalent formulations. To ease the notations in your question, since here $p\nmid [\mathbf Q(\mu_p):\mathbf Q]$, we may as well sum up the isotypical components relative to even (odd) characters and deal only with the $\pm1$ components of the modules involved.
(MC$1$) Sticking to your notations, the MC reads char(${C_{\infty}}^+)$ = char (${E_{\infty}}^+/{V_{\infty}}^+)$, and you ask whether this could be extended to the minus part. No, because the minus parts of the units at finite levels are reduced to $\pm 1$, so upstairs char(${C_{\infty}}^-)$ would be trivial, whereas downstairs the arithmetic of the minus part of the $p$-class group is definitely not (Kummer, Bernoulli numbers, etc.) On the opposite, the plus part is conjectured to be trivial : downstairs, it is Vandiver's conjecture that $p\nmid h^+$; upstairs, Greenberg's conjecture that ${C_{\infty}}^+$ is finite. To have an idea of the strength of these conjectures, note that Greenberg's (the weakest one) easily implies directly the MC. Actually (MC$1$), which avoids the appeal to $p$-adic $L$-functions, is rather specific of the proofs of the MC based on Euler systems.
(MC$2$) In view of the comments above concerning (MC$1$), it appears that - at the time being - our true knowledge of the arithmetic of the $p$-class groups is rather one-handed. To stress the most striking feature, the link with $p$-adic $L$-functions, we must come back to characters, since the trivial character must be excluded because of the pole at $s=1$. The character-wise formulation (MC$2$) on the minus size then reads : Let $\omega$ denote the Teichmüller character. For any non trivial even character $\chi$ of Gal($\mathbf Q(\mu_p)/\mathbf Q)$, the characteristic series $f_{\chi}$ of the $\omega \chi^{-1}$-part of ${C_{\infty}}$ satisfies $f_{\chi}((1+p)^s -1)=L_p(\chi, s)$ for all $s\in \mathbf Z_p$ .
(MC$3$) To come back to the plus size we must change modules, introducing ${B_{\infty}}$= the Galois group over $\mathbf Q(\mu_{p^\infty})$ of the maximal pro-p-abelian extension of $\mathbf Q(\mu_{p^\infty})$ which is $p$-ramified, i.e. unramified outside $p$. Then (MC$3$) reads : For any non trivial even character $\chi$ of Gal($\mathbf Q(\mu_p)/\mathbf Q)$, the characteristic series $g_{\chi}$ of the $ \chi$-part of ${B_{\infty}}$ satisfies $f_{\chi}((1+p)^{1-s} -1)=L_p(\chi, s)$ for all $s\in \mathbf Z_p$ . Comparing with (MC$2$), note the shift between the pair ($s,\omega \chi^{-1}$) and the pair ($1-s, \chi^{-1}$), which is due to a combination of class-field theoretic iso. and Kummer duality, usually called Spiegelung ./.
Best Answer
The action is trivial on $\Bbb Z_\ell(0)$. There is a Galois action on the $\mu_{\ell^n}$: each $\sigma$ in the Galois group maps each element of $\mu_{\ell^n}$ to some element of $\mu_{\ell^n}$. This induces the action on $\Bbb Z_\ell(1)$ (so it's basically via the Galois group of the part of the cyclotomic extension generated by $\ell$-power roots of unity).
For general $\Bbb Z_l(n)$ the action comes from the usual way of defining tensor products of $\Bbb G$-modules as $\Bbb G$-modules, etc.