First of all, this sum equals $-1$, not $1$.
As you have already noticed:
\begin{align}
\sum_{k=1}^{3n-1} (-1)^k \cos\left(\frac{k\pi}{3n}\right)^n &= \sum_{k=1}^{3n-1} (-1)^k\left(\frac 12\left(\exp\left(i\frac{k\pi}{3n}\right)+\exp\left(-i\frac{k\pi}{3n}\right)\right)\right)^n \\
&= 2^{-n} \sum_{k=1}^{3n-1} (-1)^k \sum_{j=0}^n \binom nj \exp\left(i\frac{(2j-n)k\pi}{3n}\right)
\end{align}
Since $\exp(kx)=\exp(x)^k$, one gets a sum of truncated geometric series. Since $0<|(j+n)/(3n)|<1$ and thus $\exp\left(2\pi i\frac{(j+n)}{3n}\right)\neq 1$ for $0\le j\le n$, this can be rewritten as follows:
\begin{align}
2^{-n} \sum_{k=1}^{3n-1} (-1)^k \sum_{j=0}^n \binom nj \exp\left(i\frac{(2j-n)k\pi}{3n}\right) &= 2^{-n} \sum_{j=0}^n \binom nj \sum_{k=1}^{3n-1} \left(-\exp\left(i\frac{(2j-n)\pi}{3n}\right)\right)^k \\
&= 2^{-n} \sum_{j=0}^n \binom nj \sum_{k=1}^{3n-1} \exp\left(2\pi i\cdot\frac{j+n}{3n}\right)^k \\
&= 2^{-n} \sum_{j=0}^n \binom nj \frac{\exp\left(2\pi i\cdot\frac{j+n}{3n}\right)-\exp\left(2\pi i\cdot\frac{j+n}{3n}\right)^{3n}}{1-\exp\left(2\pi i\cdot\frac{j+n}{3n}\right)}\\
&= 2^{-n} \sum_{j=0}^n \binom nj \cdot(-1) \\
&= -1
\end{align}
