Let $G$ be a finite $p$-group, $k$ a field of characteristic $p$ and let $I(G)=(g-1\mid g \in G)$ be the augmentation ideal of the group ring $k[G]$. It's known that $I(G)$ is nilpotent, i.e. there is $n> 0$ such that $I(G)^n=0$. Call the least such $n$ the nilpotency degree of $I(G)$ and denote it by $\operatorname{nildeg}I(G)$.
Question 1: Are there known upper and lower bounds for $\operatorname{nildeg}I(G)$ ?
Question 2: Does the invariant $\operatorname{nildeg}I(G)$ have a particular name in the literature ?
An inspection of the proof that $I(G)$ is nilpotent can be used to detemine upper bounds for $\operatorname{nildeg}I(G)$: Let $C \le G$ be central. Then $$\frac{k[G]}{I(C)k[G]} \cong k[G/C]\;,\qquad \frac{I(G)}{I(C)k[G]} \cong I(G/C).$$ By taking $C=\mathbb{Z}/p$ and iterating, one obtains $$\operatorname{nildeg}I(G)\le |G|.$$ By taking $C=Z(G)$ this can be futher refined: If $1 = Z_0 \le Z_1 \le ... \le Z_c = G$ is the upper central series of $G$ and $Z_i/Z_{i-1}=\prod_{j=1}^{r_i}\mathbb{Z}/p^{e_{ij}}$ then $$\operatorname{nildeg}I(G) \le \prod_{i=1}^c\;\big((e_{i,1}-1) + \cdots + (e_{i,r_i}-1)+1\big)$$ Since $(g-1)^{\operatorname{nildeg}I(G)}=0$ for each $g \in G$, a trivial lower bound is $$\operatorname{nildeg}I(G) \ge \operatorname{exp}(G)/p$$ Hence a more acurate formulation for quest 1 is:
Question 3: Are there better bounds than these or bounds that use other invariants of $G$ ?
Edit: Apart from the exact formula given by Jennings' theorem as described in mt's answer, I found the lower bound
$$\operatorname{nildeg}I(G) \ge m(p-1)+1$$
if $|G|=p^m$ in the book "Karpilovsky: The Jacobson Radical of Group Algebras".