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

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.