If we don't insist on writing it with the $\dfrac{1+\cos\theta}{1-\cos\theta}$ form, a factorisation is easy to be had. We have
$$\sum_{k = 0}^{\left\lfloor \frac{n-1}{2}\right\rfloor} (-1)^k \binom{n}{2k+1} \frac{x^{2k}}{n^{2k+1}} = \frac{\bigl(1 + i\frac{x}{n}\bigr)^n- \bigl(1 - i\frac{x}{n}\bigr)^n}{2ix}.$$
Writing
$$1 \pm i \frac{x}{n} = \sqrt{1 + \frac{x^2}{n^2}}\cdot \exp \Biggl(\pm i\arctan \frac{x}{n}\biggr)$$
we see that
$$\frac{\bigl(1 + i\frac{x}{n}\bigr)^n- \bigl(1 - i\frac{x}{n}\bigr)^n}{2ix} = \biggl(1+\frac{x^2}{n^2}\biggr)^{n/2}\cdot \frac{\sin\bigl(n\arctan \frac{x}{n}\bigr)}{x},$$
so the zeros of the polynomial are
$$\pm n\tan \frac{k\pi}{n},\quad 1 \leqslant k \leqslant \biggl\lfloor\frac{n-1}{2}\biggr\rfloor.$$
Grouping the zeros in pairs of negatives, we obtain the factorisation
$$\sum_{k = 0}^{\left\lfloor \frac{n-1}{2}\right\rfloor} (-1)^k \binom{n}{2k+1} \frac{x^{2k}}{n^{2k+1}} = \prod_{k = 1}^{\left\lfloor \frac{n-1}{2}\right\rfloor}\biggl( 1 - \frac{x^2}{n^2\tan^2 \frac{k\pi}{n}}\biggr),$$
since the constant term of the polynomial is $1$.
I unfortunately don't see a natural way that leads to the form
$$\prod_{k = 1}^{\left\lfloor \frac{n-1}{2}\right\rfloor} \biggl( 1 - \frac{(1+\cos (2k\pi/n))x^2}{(1 - \cos (2k\pi/n))n^2}\biggr).$$
Best Answer
$$f(k) = \sum^\infty_{n=1}\frac{1}{k^n-1}=\frac{\log \left(\frac{k}{k-1}\right)-\psi _{\frac{1}{k}}^{(0)}(1)}{\log (k)}$$ where appears the q-digamma function.
For large values of $k$, series expansion gives $$f(k)=\frac{1}{k}+\frac{2}{k^2}+\frac{2}{k^3}+\frac{3}{k^4}+\frac{2}{k^5}+\frac{4}{k^6}+\frac{2}{k^7}+\frac{4}{k^8}+O\left(\frac{1}{k^9}\right)$$ For $k=4$, this would give $\frac{6899}{16384}\approx 0.421082$ which is not too bad compared to the value you gave for $f(4)$.