17. $\frac{7}{3}$.

Let $t=x+\frac{1}{x}$. Then $t \geqslant 2$,
$$\begin{array}{l} x^{4}+\frac{1}{x^{4}}=t^{4}-4 t^{2}+2, \\ x^{3}+\frac{1}{x^{3}}=t^{3}-3 t . \end{array}$$

Thus $f(x)=g(t)=\frac{4}{3} t-\frac{2}{3} t^{-1}$
$$\geqslant g(2)=\frac{4}{3} \times 2-\frac{2}{3} \times \frac{1}{2}=\frac{7}{3}=f(1) \text {. }$$

Therefore, the minimum value of $f(x)$ is $\frac{7}{3}$.