The largest permutation group without 2-cycles is $A\_n$, which has size $n!/2$. I think the largest permutation group without 2-cycles or 3-cycles is much smaller, but I can't figure out if it should be polynomially smaller (eg. of size $n!/n^3$), or more dramatically smaller (eg. of size $(.5n)!$). 


The largest group I could come up with is $\{\phi(x\_1,\dots,x\_{n/2}) \circ \phi(x\_{n/2+1},\dots,x\_n) | \phi \in S\_{n/2}\}$, which has size $(.5n)!$.




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EDIT: Posting this here since the answers below pointed me in the right direction, but ended 
up conjecturing something that was not quite correct. The group 


$\{(g\_1, g\_2, \dots, g\_{d-1}\\!,\ g\_1g\_2\cdots g\_{d-1}\\!)\ |\ g\_i \in S\_{n/d}\} \le S\_{n/d}^d \le S\_n$


has no 2-cycles or 3-cycles, and has $(n/d)!^{d-1}$ elements. When $d = \log n$, this is $n!/n^{\Theta(n\log\log(n)/\log(n))}$, which is smaller than $n!/poly(n)$ but larger than $(cn)!$ for any $c<1$. 


You can do a little bit better by using a wreath product instead of a direct product, and by tweaking $d$, but I think this is more or less optimal.



Largest permutation group without 2-cycles or 3-cycles
I can do a bit better! For even $n=2m$ there is a subgroup of order $2^{m-1}m!$ with no 2-cycles or 3-cycles. Let $W$ be the wreath product of a cyclic group of order 2 with $S\_m$. In other words, $W$ is the subgroup of $S\_n$ that preserves the partition $\{\{1,2\},
\{3,4\},\ldots,\{2m-1,2m\}\}$ of $\{1,2,\ldots,n\}$. So $|W| = 2^mm!$.


Then $W$ has no 3-cycles, but it does have the 2-cycles $(1,2),(3,4),$ etc. Now $W$ is a semidirect product $B \rtimes S\_m$, where $B$ is the base group of the wreath product, which is the subgroup of order $2^m$ that fixes each of the pairs $\{2i-1,2i\}$ in the partition. The subgroup $C$ of $B$ consisting of the even permutations in $B$ has order $2^{m-1}$ and has no 2-cycles, and it is normal in $W$, so $G := C \rtimes S\_m$ has no 3-cycles and no 2-cycles.


I would guess that this is the best you can do for large $n$, and I am sure that this could be proved using the methods suggested by Dima Pasechnik. The intransitive and imprimitive maximal subgroups of $S\_n$ are respectively direct and wreath products of symmetric groups, so their structure is well understood. The primitive maximal subgroups are comparatively small. There is an old result (of Praeger and Saxl I think) that says they have order at most $4^n$, and many more recent more accurate results, but $4^n$ is already asymptotically smaller than $|G|$.


