Right. What I'm looking for should be something like an idempotented algebra (it is supposed to admit a map from the Hecke algebra of a p-adic group)

No, I don't think that's it: in order to be able to talk about a $\mathfrak{C}$-module as a $k$-module, you need some sort of fiber functor $\mathfrak{C}\to \text{vect}$, which I'm assuming isn't given.

@DavidLoeffler Thanks! Localizing the coefficient field at a prime is certainly ok. I'm not very literate in Galois representations formalism, but the Poitou-Tate result in Nekovar's notes (0.7.1) looks very applicable. Is this what you meant?

(with restricted product above). As maps to $\mathbb{Q}/\mathbb{Z}$ is an exact contravariant functor and we're hoping $H^0(K, \mathbb{G}_m)$ is Galois-invariant points of $\bar{K}^\times$ (or $\pi_0 K^\times$ in the infinite place case), the exactness of the sequence above would imply this CFT statement.

Sure. It's that a Galois character is equivalent to a map from $\prod_{v} K_v^\times$ to $\mathbb{Q}/\mathbb{Z}$ which vanishes on $K^\times.$

The product map $S_n\times S_m\to S_{m+n}$ should give a coproduct on this cohomology, and you're looking for a class $\alpha$ whose coproduct is $\alpha\otimes 1 + 1\otimes \alpha.$ Maybe this helps narrow it down?

In this particular case, since Abelian varieties are a full subcategory of pointed varieties, I think you can apply part 5 of Definition 2.1 here ncatlab.org/nlab/show/idempotent+monad to show that indeed taking the Albanese variety is monadic.

Note that any pair of adjoint functors $L:C\leftrightarrows D :R$ give rise to a monad via the natural transformations $RLRL\overset{LR\to 1}{\to} RL$ and $1\to RL$ given by the unit/counit maps of the adjunction. Any object of $D$ then gives an algebra over this monad via the transformation $RLR\overset{LR\to 1}{\to} R$. A pair is said to be "monadic" if in fact $D$ is equivalent to the category of algebras over $C$. This condition can be checked formally (and holds for most "free-like" functors): see en.wikipedia.org/wiki/Beck%27s_monadicity_theorem.