Solution:

Answer: $n=1$ is clearly a solution, we can just color each side of the equilateral triangle in a different color, and the conditions are satisfied. We prove there is no larger $n$ that fulfills the requirements.

Lemma 1. Given a regular $2n+1$-gon in the plane, and a sequence of $n+1$ consecutive sides $s_{1}, s_{2}, \ldots, s_{n+1}$ there is an external point $Q$ in the plane, such that the color of each $s_{i}$ can be seen from $Q$, for $i=1,2, \ldots, n+1$.

Proof. It is obvious that for a semi-circle $S$, there is a point $R$ in the plane far enough on the perpendicular bisector of the diameter of $S$ such that almost the entire semi-circle can be seen from $R$.

Now, it is clear that looking at the circumscribed circle around the $2n+1$-gon, there is a semi-circle $S$ such that each $s_{i}$ either has both endpoints on it, or has an endpoint that is on the semi-circle, and is not on the semicircle's end. So, take $Q$ to be a point in the plane from which almost all of $S$ can be seen, clearly, the color of each $s_{i}$ can be seen from $Q$. $\diamond$

Take $n \geq 2$, denote the sides $a_{1}, a_{2}, \ldots, a_{2n+1}$ in that order, and suppose we have a coloring that satisfies the condition of the problem. Let's call the 3 colors red, green and blue. We must have 2 adjacent sides of different colors, say $a_{1}$ is red and $a_{2}$ is green. Then, by Lemma 1:

(i) We cannot have a blue side among $a_{1}, a_{2}, \ldots, a_{n+1}$.

(ii) We cannot have a blue side among $a_{2}, a_{1}, a_{2n+1}, \ldots, a_{n+3}$.

We are required to have at least one blue side, and according to (i) and (ii), that can only be $a_{n+2}$, so $a_{n+2}$ is blue.

Now, applying Lemma 1 on the sequence of sides $a_{2}, a_{3}, \ldots, a_{n+2}$ we get that $a_{2}, a_{3}, \ldots, a_{n+1}$ are all green.

Applying Lemma 1 on the sequence of sides $a_{1}, a_{2n+1}, a_{2n}, \ldots, a_{n+2}$ we get that $a_{2n+1}, a_{2n}, \ldots, a_{n+3}$ are all red.

Therefore $a_{n+1}, a_{n+2}$ and $a_{n+3}$ are all of different colors, and for $n \geq 2$ they can all be seen from the same point according to Lemma 1, so we have a contradiction.