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.


It is even true for $n\ge3$. Indeed $\;\dbinom 93=\dfrac{9!}{3!\,6!}=84 >64$.
\begin{align}
&\text{Next, we have }\qquad\qquad
&\binom{3(n+1)}{n+1}&=\binom{3n}n\,\frac{(3n+1)(3n+2)(3n+3)}{(n+1)(2n+1)(2n+2)},&&\qquad\qquad\end{align}
so, by induction, all we have to prove is that
$$\frac{3(3n+1)(3n+2)}{2(n+1)(2n+1)}\ge4$$
This inequality is equivalent to
$$ 27n^2+27n+6\ge 16n^2+24n+8\iff 11n^2+3n\ge 2, $$
which is obviously true if $n \ge 1$.
