For $n \ge 4$, does it follow that ${{3n} \choose {n}} > 4^n$ I believe that the answer is yes.  Here's my thinking:
(1) For $n \ge 4, { {2n} \choose {n}} \ge \frac{4^n}{n}$

By induction:  $64 = \frac{4^4}{4} \le {{8}\choose{4}}= 70$. Assume it is true up to $n-1 \ge 4$.  Then, ${ {2n} \choose n} = 2\left(\frac{2n-1}{n}\right){{2n-1}\choose{n-1}} > 2\left(\frac{2n-1}{n}\right)\frac{4^{n-1}}{n-1} > 2 \cdot 2\cdot \frac{4^{n-1}}{n} = \frac{4^n}{n}$
Note:  This argument was taken from Wikipedia.

(2)  $(3n-1) > \left(\frac{3}{2}\right)(2n-1), (3n-2) > \left(\frac{3}{2}\right)(2n-2), \dots, (2n-n+1) > \left(\frac{3}{2}\right)(2n-n+1)$

If $2a = 3b$, then $2a - 2 > 3b - 3$ and $2(a-1) > 3(b-1)$ so that $(a-1) > \frac{3}{2}(b-1)$

(3)  For $n\ge 2$, ${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n} \choose {n}}$

From (2) above: ${{3n} \choose{n}} = \frac{(3n)(3n-1)\dots(3n-n+2)(3n-n+1)}{n!} > \left(\frac{3}{2}^n\right)\frac{(2n)(2n-1)\dots(2n-n+2)(2n-n+1)}{n!} = \left(\frac{3}{2}^n\right){{2n} \choose {n}}$

(4)  Since for $n \ge 4, \left(\frac{3}{2}\right)^n > n$, it follows that:

${{3n}\choose{n}} > \left(\frac{3}{2}\right)^n{{2n}\choose{n}} > n\left(\frac{4^n}{n}\right) = 4^n $
Note: The argument that for $n \ge 2, \left(\frac{3}{2}\right)^n > n$ can be found here.


Your thinking seems correct to me. Here's a similar approach: The cases $n=4,5$ can be checked by hand. For $n\geq6$ we have $n!\leq\left(\frac{n}{2}\right)^n$, which is easily proved by induction, and hence
$$\binom{3n}{n}=\frac{(3n)!}{(2n)!n!}=\prod_{i=k}^n\frac{2n+k}{k}=\prod_{k=1}^n\left(1+\frac{2n}{k}\right)>2^n\frac{n^n}{n!}\geq4^n,$$
where the first inequality comes from the fact that $1+\frac{2n}{k}>\frac{2n}{k}$ for all $k$.
