(You might be interested in the 'offset modular generalization' in my answer here. Also see modular arithmetic and root of unity for preliminary material from Wikipedia. I will use $\lfloor\cdot\rfloor$ as the floor function and $a\equiv b(n)$ as shorthand for $a\equiv b\bmod n$.)
Let $\omega=\exp(2\pi i/3)$ be our choice of primitive third root of unity. We'll work with
$f(n)=\sum_{k=0}^{\lfloor n/3\rfloor}\binom{n}{3k+1}=\sum_{m\equiv1(3)}\binom{n}{m}=\sum_m\binom{n}{m}[m\equiv1(3)].$
Recall that if $m$ is an integer outside of the range $0\le\circ\le n$, the coefficient $\binom{n}{m}$ is zero, so the above sum makes sense. With the binomial theorem and the fact that $1+\omega+\omega^2=0$, we can rewrite the Iverson bracket $[m\equiv1(3)]$ algebraically as $(1+\omega^{m-1}+\omega^{2(m-1)})/3$ (this expression has period $3$ as a function of $m$, so to check this it suffices to check $m=0,1,2$; for motivation behind this see my linked answer). Hence
$\begin{array}{c l} f(n) & =\sum_{m=0}^n \binom{n}{m}\frac{1+\omega^{m-1}+\omega^{2m-2}}{3} \\ & =\frac{1}{3}\sum_{m=0}^n \binom{n}{m}+\frac{\omega^{-1}}{3}\sum_{m=0}^n \binom{n}{m}\omega^m+\frac{\omega^{-2}}{3}\sum_{m=0}^n \binom{n}{m}\omega^{2m} \\ & = \frac{(1+1)^n+\omega^{-1}(1+\omega)^n+\omega^{-2}(1+\omega^2)^n}{3} \end{array}$
As WimC notes, $\lambda=1+\omega$ is a sixth root of unity with conjugate $1+\omega^{-1}$ (remember $\omega^{-1}=\omega^2$ as well), so by triangle inequality
$|3f(n)-2^n|\le |\omega^{-1}(1+\omega)^n|+|\omega^{-2}(1+\omega^2)^n|\le 1+1=2.$
Hence the error between $f(n)$ and $2^n/3$ oscillates indefinitely, but will always be bounded in magnitude by $2/3$.