Yes, here's an elementary argument. $\frac{(3n)!}{n!^3}$ is the multinomial coefficient ${3n \choose n, n, n}$, so it appears in the multinomial expansion
$$(x + y + z)^{3n} = \sum_{a+b+c=3n} {3n \choose a, b, c} x^a y^b z^c$$
and plugging in $x = y = z = 1$ gives the upper bound
$${3n \choose n, n, n} \le 3^{3n}.$$
On the other hand, we can show by elementary but slightly tedious arguments (computing the ratios between consecutive coefficients) that ${3n \choose n, n, n}$ is actually the largest coefficient in this multinomial expansion (the same is true for ${2n \choose n}$ and the binomial expansion). The total number of terms in this expansion is quadratic in $n$; we don't need the exact count but it's easy to see that there are at most $(3n)^2$ terms. So the entire sum is at most $(3n)^2 {3n \choose n, n, n}$, which gives a lower bound matching our upper bound
$$\frac{3^{3n-2}}{n^2} \le {3n \choose n, n, n} \le 3^{3n}.$$
Now use the squeeze theorem.
In general people overrely a lot on Stirling's approximation when it isn't necessary, and sometimes using it is actually much more tedious and annoying than alternative elementary arguments.
Edit: Here is the elementary but slightly tedious argument I alluded to above that ${3n \choose n, n, n}$ is the largest trinomial coefficient. We just have to compute the effect of moving around the $3n$-th layer of Pascal's pyramid, which we can do by taking ratios of factorials, e.g.
$$\frac{ {3n \choose a, b+1, c-1} }{ {3n \choose a, b, c} } = \frac{b! c!}{(b+1)! (c-1)!} = \frac{c}{b+1}$$
and similarly for movements in the other directions. The result we get, although it's a little hard to see when written this way, is that moving in a given direction increases the size of the trinomial coefficient iff we are moving towards the center of Pascal's pyramid, and otherwise decreases it; equivalently, iff the three indices $a, b, c$ are getting closer. So the largest coefficient occurs at the center of the pyramid, or equivalently when the three indices are as close as possible.
Possibly a more conceptual argument, which hopefully should be believable from just staring at Pascal's pyramid a bit, is to argue by induction that the trinomial coefficients are unimodal in the above sense (bigger towards the center), and show that if the $n$-th layer of the pyramid satisfies this property then so does the $n+1$-th layer, by applying Pascal's rule. It's much more annoying to explain what this means verbally than to see it visually. This should all also be intuitively appealing from thinking in terms of outcomes from rolling a $3$-sided die; ${3n \choose n, n, n}$ corresponds to rolling the same number of each side which should intuitively be the most likely outcome.
