If $$f(x)=\prod_{k=1}^\infty\left(1-\frac{x^3}{k^3\pi^3}\right)$$ and if $$f(x)=\sum_{n=0}^\infty a_n x^n$$ Then find $a_n$ or convert the above infinite product into an infinite sum.
We see that $f(k\pi)=0$ where $k\in\mathbb{N}$
Do we need logarithms?
On
An "algorithmic" way is via doing the exponentiation in $$f(x)=\exp\left[-\sum_{k=1}^\infty\sum_{n=1}^\infty\frac1n\left(\frac{x}{k\pi}\right)^{3n}\right]=\exp\left[-\sum_{n=1}^\infty\frac1n\frac{\zeta(3n)}{\pi^{3n}}x^{3n}\right].$$ Some "general" closed forms may then be taken from the linked Wiki page. I wouldn't hope for anything more specific, because of the (conjectural) independence of the "odd $\zeta$-values".
Thus, let $f(x)=\sum_{n=0}^\infty (-1)^n c_n x^{3n}$ and $r_n=\zeta(3n)/\pi^{3n}$; then the first few values of $c_n$ are \begin{align} c_0&=1,\\ c_1&=r_1,\\ c_2&=\frac12(r_1^2-r_2),\\ c_3&=\frac16(r_1^3-3r_1r_2+2r_3),\\ c_4&=\frac1{24}(r_1^4-6r_1^2r_2+8r_1r_3+3r_2^2-6r_4),\\ c_5&=\frac1{120}(r_1^5-10r_1^3r_2+20r_1^2r_3+15r_1r_2^2-30r_1r_4-20r_2r_3+24r_5), \end{align} and so on, and we can put in the known $r_2=1/945$, $r_4=691/638512875$, etc.
On
Let $a=\frac x \pi$
$$f(a)=\prod_{k=1}^\infty\left(1-\frac{a^3}{k^3}\right)=\frac{\prod_{k=1}^\infty(k-a)(k-b)(k-c)}{\prod_{k=1}^\infty k^3}=\frac{\prod_{k=1}^\infty(k-a)}{\prod_{k=1}^\infty k}\frac{\prod_{k=1}^\infty(k-b)}{\prod_{k=1}^\infty k}\frac{\prod_{k=1}^\infty(k-c)}{\prod_{k=1}^\infty k}$$
where $$b=-\frac{a}{2} \left(1+i \sqrt{3}\right)\qquad \text{and} \qquad c=-\frac{a}{2} \left(1-i \sqrt{3}\right)$$ Using Pochhammer symbols $$\frac{\prod_{k=1}^p(k-d)}{\prod_{k=1}^p k}=\frac{(1-d)_p}{p!}$$ taking the limit and then using the gamma function $$f(a)=\frac{1}{\Gamma (1-a) \Gamma (1-b) \Gamma (1-c)}$$ Replacing $b$ and $c$ by their values it is easy to compute $$\frac 1{f(a)}=1+\sum_{i=1}^\infty \frac{C_n}{(3n)!} a^{3n}$$ where the first $C_n$ are given in terms of $\zeta$ functions $$\left( \begin{array}{cc} n & C_n \\ 1 & 6 \zeta (3) \\ 2 & \frac{8}{21} \left(\pi ^6+945 \zeta (3)^2\right) \\ 3 & 192 \left(\pi ^6 \zeta (3)+315 \left(\zeta (3)^3+2 \zeta (9)\right)\right) \\ 4 & \frac{89472 \pi ^{12}}{455}+126720 \pi ^6 \zeta (3)^2+19958400 \left(\zeta (3)^4+8 \zeta (3) \zeta (9)\right)\\ 5 & 2304 \left(233 \pi ^{12} \zeta (3)+50050 \pi ^6 \left(\zeta (3)^3+2 \zeta (9)\right)+4729725 \left(\zeta (3)^5+20 \zeta (3)^2 \zeta (9)+24 \zeta (15)\right)\right) \end{array} \right)$$
Where the problem starts to be tedious is when we nedd to perform the long division to obtain $f(a)$ itself.
A much nicer way is to write $$\log(f(a))=-\sum_{n=1}^\infty \frac{\zeta (3 n)}{n}\,a^{3n}$$ Then, use $$f(a)=e^{\log(f(a)}$$ and use Taylor series for nasty coefficients as above to have the polynomial.
If we let $\alpha_n=-\frac{\zeta (3 n)}{n}$ and $y=a^3$ $$f(a)=\exp\Bigg[\sum_{n=1}^\infty \alpha_n\,y^{n}\Bigg]=1+\sum_{n=1}^\infty \frac{\beta_n}{n!}\,y^{n}$$ and the first $\beta_n$ are given by
$$\left( \begin{array}{cc} n & \beta_n \\ 1 & \alpha_1 \\ 2 & \alpha_1^2+2 \alpha_2 \\ 3 & \alpha_1^3+6 \alpha_1 \alpha_2+6 \alpha_3 \\ 4 & \alpha_1^4+12 \alpha_1^2\alpha_2+24 \alpha_1 \alpha_3+12 \alpha_2^2+24 \alpha_4 \\ 5 & \alpha_1^5+20 \alpha_1^3\alpha_2+60 \alpha_1^2\alpha_3+60 \alpha_1\alpha _2^2 +120 \alpha_1\alpha_4+120 \alpha_2 \alpha_3+120 \alpha_5 \end{array} \right)$$
remark. If $$ \prod_{k=1}^\infty\left(1-\frac{x^3}{k^3\pi^3}\right) =\sum_{n=0}^\infty a_n x^n $$ then \begin{align} a_0 &= 1 ;\\ a_3 &= \sum_{k=1}^\infty \frac{-1}{k^3\pi^3} = -\frac{\zeta(3)}{\pi^3} \approx -0.03876817958 ;\\ a_6 &= \sum_{1\le k < m}\frac{1}{k^3m^3\pi^6} \approx 0.0002223853458 . \end{align} So perhaps the first thing to do is "closed form for $a_6$", if any.
$$ a_{3n} = \frac{(-1)^n}{\pi^{3n}}\prod\frac{1}{(k_1k_2\dots k_n)^3} , $$ where the product is over all $n$-tuples $(k_1,k_2,\dots,k_n)$ with $1 \le k_1<k_2<\cdots < k_n$.
Graph of $f(x)$
