Does there exist an $n$ such that all groups of order $n$ are Abelian?

I know that all groups of order $\leq$ 5 are Abelian and all groups of prime order are Abelian. Are there any other examples? If so is there something special about the orders of these groups?

Yes, such numbers are called abelian numbers.

A number $n$ is abelian, if and only if it is cube-free and there is no prime power $p^k\mid n$ with $k\ge 1$, such that $p^k\equiv1\pmod q$ for some prime $q\mid n$.

In particular, if $p$ is a prime, then $p$ and $p^2$ are abelian. The number $pq$ with $2<p<q$ is abelian, if and only if $p$ does not divide $q-1$. The only even abelian numbers are $2$ and $4$.

The terminology you're looking for is:

Definition. A number $n \in \mathbb{N}$ is said to be

• cyclic iff all groups of order $n$ are cyclic.

• abelian iff all groups of order $n$ are abelian.

• nilpotent iff all groups of order $n$ are nilpotent.

Using basic group theory, we see that these are related as follows:

Proposition. For natural numbers, we have: $$\mbox{ prime } \rightarrow \mbox{ cyclic } \rightarrow \mbox{ abelian } \rightarrow \mbox{ nilpotent }$$

A Peter explains, your question is essentially: does there exist an abelian number strictly greater than $5$ that isn't prime? Yes, $15$ does the trick. In fact, $15$ has the stronger property of being a cyclic number strictly greater than $5$ that isn't prime.

You can have a look at this mathoverflow post for more information.