Fix some $n \geq 1$ and some prime $p$. I'm looking for finite $p$-groups $G$ and finite-dimensional complex representations $V$ of $G$ with the following two properties:


1. The abelianization of $G$ has rank $n$; for instance, it could be $(\mathbb{Z}/p)^n$.
2. For all nonidentity $g \in G$ and all nonzero $v \in V$, we have $g(v) \neq v$. In other words, $1$ is not an eigenvalue for the action of $g$ on $V$.


Here are the cases I know how to deal with:


1. For $n=1$ and an arbitrary prime $p$, you can take the cyclic group of order $p$ and the $1$-dimensional representation where the generator acts as rotation by $2\pi/p$.
2. For $n=2$ and $p=2$, you can let $G$ be the finite quaternion group and $V$ be the unique $2$-dimensional irreducible representation of $G$.



Representations of p-groups where 1 is never an eigenvalue
(This question has been answered in comments by Geoff Robinson:) 


You have already found (almost) all examples there are, namely cyclic $p$-groups and generalized quaternion groups in case $p=2$. The point is that every abelian subgroup of such a group $G$ is cyclic: Suppose $A\leq G$ is abelian. Let $\lambda \in \operatorname{Irr}(A)$ be a linear constituent of the character of $V$. Then every element in $\ker(\lambda)$ has non-zero fixed vectors in $V$. Thus $\ker(\lambda)=1$ and $A$ is cyclic. 


By (say) Theorem 6.11 in Isaacs' text *Finite Group Theory*, a $p$-group with all abelian subgroups cyclic is cyclic or generalized quaternion. (This result should be in many other books on finite group theory.) 


Finite groups ($p$-groups or not) which admit a complex representation such that Condition 2. holds are exactly the Frobenius complements ([see here](https://mathoverflow.net/q/191769/10266)).


