I see 'self injective dimension' more often, I think. But yes, it is just the injective dimension of the ring as a module over itself.

$R$ (left) Noetherian is both necessary and sufficient: if $R$ has a non-f.g. left ideal $I$ then the kernel of $R\to R/I$ will be $I$ and so not be finitely generated. The converse is not difficult.

$G$-graded ring? en.wikipedia.org/wiki/…

Isn't the trouble with this products of two fields example that you have to ensure the fields don't have the same characteristic? I suppose you could add the condition that there is no field of that order.

Any book on the representation theory of Lie algebras would persuade you that the answer is no in general.

What do you mean by a simple Noetherian commutative ring? A field?

Sorry. I just noticed that I typed polynomial when I meant power series.

In the case you just added the ring is isomorphic to the formal polynomial ring in one variable $k[[T]]$. The element $T$ corresponds to an element $g-1$ with $g$ a generator of $\mathbb{Z}_p$. The references you have already been given prove this.

Perhaps more accurately the notation for the complete group algebra is sometimes k[[G]]. In the case the OP is interested in I have also seen $[[kG]]$, $\Omega(G)$ or $\Omega_G$. As it is the most natural algebraic object to attach to the pair $(k,G)$ I would normally use $kG$ myself.