For one triangle there are actually $n^3-n$ edge colourings, including reflections and rotations.
There are $6$ edges in a tetrahedron, $n^6$ ways to colour the edges, $4$ triangles, $n^4$ colourings with any given triangle monocromatic, $6$ pairs of triangles, $n^2$ colourings with any given pair of triangles both monochromatic, 4 ways to pick $3$ triangles, $n$ colourings with $3$ triangles, hence all triangles, monochromatic, $1$ way to pick $4$ triangles and $n$ colourings with $4$ triangles monochromatic.
The number of $n$-colourings without monochromatic triangles is then
$n^6-4n^4+6n^2-3n$.
You can simplify(?) your argument by using induction:
Let $f(k)=R(\underbrace{3,3,\ldots,3}_k)$.
In any $k$-colouring ($k\ge1$) of a $K_n$ graph ($n\ge2$), if we pick any vertex $v$, then by the pigeon-hole-principle, one of the $k$ colours ("blue", say) will occur for at least $m:=\left\lceil\frac{n-1}{k}\right\rceil$ times among the edges incident with $v$. If any edge between the other ends of these edges is blue, we have a blue triangle and are done. And if none of these edges is blue, we have found a subgraph that is a $(k-1)$-coloured $K_m$. Hence if $m\ge f(k-1)$ we are done again.
In other words, we almost immediately find a monochromatic triangle if $\frac {n-1}k> f(k-1)-1$, or equivalently if $n>kf(k-1)-k+1$. We conclude that $$f(k)\le kf(k-1)-k+2.$$
Clearly, $f(1)=3$. From this, $f(2)\le 6$, $f(3)\le 17$, $f(4)\le 66$, etc.
Thus we have found an upper bound
$$ f(k)\le a_k$$
where $a_k$ is defined by the recursion $$\begin{align}a_1&=3,\\ a_{k}&=ka_{k-1}-k+2\qquad\text{for }k\ge2.\end{align}$$
Claim. We have $a_k-1=\sum_{i=0}^k\frac{k!}{i!}$.
Proof.
This is clear for $k=1$.
For $k\ge 2$ we find by induction
$$a_k-1=k(a_{k-1}-1)+1=k\sum_{i=0}^{k-1}\frac{(k-1)!}{i!}+\frac{k!}{k!}=\sum_{i=0}^k\frac{k!}{i!}$$
$\square$
Corollary.
$$R(\underbrace{3,3,\ldots,3}_k)\le \left\lceil k!e\right\rceil$$
Proof. This follows from $e=\sum_{i=0}^\infty\frac1{i!}$ and that the error $\sum_{i=k+1}^\infty\frac1{i!}$ is between $0$ and $1$ for $k\ge1$. $\square$
Remark. See also A073591.
Best Answer
This answer follows @Nishant's suggestion to use Burnside's Lemma.
The rotation group of the tetrahedron contains eight rotations of $\pm120^\circ.$ For a coloring to be fixed under one of these, the face through which the rotation axis passes must be monochromatic, and the three edges not on that face must also be of a single color. So these rotations fix $n^2$ colorings.
The group contains three rotations of $180^\circ$ about an axis joining the centers of a pair of opposite edges. Call these edges $e$ and $e'.$ One of these edges, say $e,$ must be incident on a monochromatic triangle. For the coloring to be fixed under rotation about the axis joining $e$ and $e',$ both triangles incident on $e$ must be monochromatic, so the five edges other than $e'$ must have the same color. This gives $n^2$ colorings fixed by the rotation. Multiplying by $2$ (because of the choice of $e$ or $e'$) and subtracting $n$ (so as not to double count the colorings where all six edges have the same color) gives $2n^2-n$ colorings fixed by $180^\circ$ rotation.
All colorings are fixed by the identity (and you know how many of these there are). Now apply Burnside's Lemma.