Use the Taylor series of $\sqrt[3]{1+x}$ to show that the term inside the floor bracket is \[8n+4-\frac{50}{3n}+\mathcal{O}\left(\frac{1}{n^2}\right).\]Hence for all sufficiently large $n$, we have $a_n=8n+3$. The claim now follows since \[(8n+3)^{8n+3} \equiv 3^{8n+3} \equiv 27 \cdot 3^{8n} \equiv 3 \cdot 9^{4n} \equiv 3 \mod 8.\]