It should be $\alpha\ge 3\sqrt{3}/2$, but a heuristic approach. Thou, it could be made rigorous, I suppose.The min $\alpha$ is attained when $\frac{x_{n+1}}{x_n}=\sqrt{3}$. The idea is to consider the min possible $\alpha$. By some continuity argument there is such one. The corresponding sequence $x_n$ should have the following property. Put $t$ instead of $x_n$ and run it trough $[x_{n-1}, x_{n+1}]$. The minimum of this function, depending on $t$, should be attained exactly when $t=x_n$. It doesn't give unique sequence $x_n$ but among all sequences this condition generates, it seems, the appointed one is the best.

Here is one way to make it "rigorous". Consider two cases: (1) $\{ x_n \}$ is bounded. Say $x_n \le M$ for all $n \in \mathbb{N}$. Thus $\frac{x_{n+1}}{x_n^3} \ge \frac{1}{x_n^2} \ge \frac{1}{M^2}$, so $\sum_{n=0}^{\infty}\frac{x_{n+1}}{x_n^3}$ is divergent, contradiction. (2) $\{ x_n \}$ is unbounded, since it is increasing, $\lim_{n \to \infty} x_n=\infty$. we have the following estimate $$ \frac{x_{n+1}}{x_n^3} \ge \frac{3\sqrt{3}}{2} \Big( \frac{1}{x_n^2}-\frac{1}{x_{n+1}^2} \Big)$$ It's equivalent to $(x_{n+1}-\sqrt{3} x_n)^2(x_{n+1}+\tfrac{\sqrt{3}}{2} x_n) \ge 0$, which is clearly true. Summing up for $n=0,1,2,\cdots$ and, as $\lim_{n \to \infty} x_n=\infty$, we have $$ \alpha=\sum_{n=0}^{\infty}\frac{x_{n+1}}{x_n^3} \ge \sum_{n=0}^{\infty} \frac{3\sqrt{3}}{2} \Big( \frac{1}{x_n^2}-\frac{1}{x_{n+1}^2} \Big) = \frac{3\sqrt{3}}{2} \frac{1}{x_0^2}=\frac{3\sqrt{3}}{2}$$ For the construction, let $\alpha \ge \frac{3\sqrt{3}}{2}$, then there is a $t \ge \sqrt{3}$ such that $t+\frac{3\sqrt{3}}{2} \cdot \frac{1}{t^2}=\alpha$. Choosing $x_n=(\sqrt{3})^{n-1} \cdot t$ for $n \ge 1$ suffices.