This is to supplement Ian's answer and get examples in all dimensions $\ge 3$. 

Let $M={\mathbb H}^n/\Gamma$ be a compact hyperbolic $n$-manifold; suppose that $f\in Homeo(M)$ is a homeomorphism of prime order $p$. Assume that $f$ is homotopic to the identity. Our goal is to reach a contradiction. Let $\tilde f: {\mathbb H}^n\to {\mathbb H}^n$ denote a lift of $f$ to the universal covering of $M$. Then, there exists $g\in \Gamma$, a deck-transformation of ${\mathbb H}^n\to M$, such that $g$ and $\tilde f$ induce (by conjugation) the same automorphism of $\Gamma$.  Extending $\tilde f$ to the boundary sphere of ${\mathbb H}^n$, we obtain that $\tilde f= g$ on $S^{n-1}$. Consider the composition $h=g^{-1}\circ \tilde f$. I claim that $h=id$, i.e. $\tilde f=g$. That would imply that $f=id_M$, contradicting our assumptions that $f$ has period $p\ge 2$. 

The map $h$ restricts to the identity on $S^{n-1}$. Since $f^p=id_M$, there exists $\gamma\in \Gamma$ such that $h^p=\gamma$. Since $h^p$ restricts to the identity on $S^{n-1}$, it follows that $\gamma=1\in \Gamma$. Thus, $h$ is periodic. I will be using the conformal model of ${\mathbb H}^n$, identified with with the open unit ball $B^n$ in ${\mathbb R}^{n}$. Thus, extend $h$ by the identity to the exterior of $B^n$ in $S^n={\mathbb R}^{n}\cup \{\infty\}$. The extension, again denoted $h$, is $p$-periodic and its fixed-point set has nonempty interior in $S^n$. Unless $h=id$, this contradicts Newman's theorem [1] or, if you prefer, the contradicts the Smith Theory [2], according to which the fixed-point set of a $p$-periodic homeomorphism of $S^n$ is either empty or is a $p$-homology manifold and $p$-homology sphere. 

The conclusion, therefore, is that if $f: M\to M$ is $p$-periodic, $p$ is prime, then $f$ cannot be homotopic to the identity, hence, by Mostow Rigidity, is homotopic to a $p$-periodic isometry of $M$. Now, one uses the existence of compact hyperbolic $n$-manifolds for all $n\ge 3$, with trivial isometry group, just as in Ian's argument. To conclude: For all $n\ge 3$ there exist smooth compact $n$-manifolds without nontrivial periodic self-homeomorphisms. 

[1] <cite authors="Newman, M. H. A.">_Newman, M. H. A._, [**A theorem on periodic transformations of spaces**](https://doi.org/10.1093/qmath/os-2.1.1-a), Q. J. Math., Oxf. Ser. 2, 1-8 (1931). [ZBL0001.22703](https://zbmath.org/?q=an:0001.22703).</cite>



[2] <cite authors="Bredon, Glen E.">_Bredon, Glen E._, [**Introduction to compact transformation groups**](https://mathoverflow.net/a/375042), Pure and Applied Mathematics, 46. New York-London: Academic Press. XIII,459 p.  (1972). [ZBL0246.57017](https://zbmath.org/?q=an:0246.57017).</cite>