If one starts with the Weierstrass factorisation
$$ \Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left(1 + \frac{z}{k}\right)^{-1} e^{z/k}$$
of the Gamma function, applied to $z = -x, -\omega x, -\omega^2 x$ (where $\omega = e^{2\pi i/3}$ is the cube root of unity), and multiplies the three together, one obtains
$$ \Gamma(-x) \Gamma(-\omega x) \Gamma(-\omega^2 x) = \frac{1}{-x^3} \prod_{k=1}^\infty \left(1-\frac{x^3}{k^3}\right)^{-1}$$
and hence
$$\prod_{k=1}^\infty \left(1-\frac{x^3}{k^3}\right) = -\frac{1}{x^3 \Gamma(-x) \Gamma(-\omega x) \Gamma(-\omega^2 x)}.$$
One can manipulate the right-hand side a little using the various functional equations of the Gamma function, but it does not have a dramatically simpler form (as one can already see from the zero set ${\bf N} \cup \omega {\bf N} \cup \omega^2 {\bf N}$, which is too weird to come from anything more elementary than a Gamma function). This is in contrast to the analogous identity
$$ \prod_{k=1}^\infty \left(1 - \frac{x^2}{k^2}\right) = -\frac{1}{x^2 \Gamma(-x) \Gamma(x)}$$
where the Euler reflection formula (and $\Gamma(1-x) = -x \Gamma(-x)$) applies to simplify the right-hand side to $\frac{\sin(\pi x)}{\pi x}$ as you mention, and the zero set ${\bf N} \cup -{\bf N} = {\bf Z} \backslash \{0\}$ is now simple enough to be explainable via trigonometric functions. (The different structure of the zero sets also helps explain why the zeta function evaluated at even natural numbers is much more tractable than the zeta function evaluated at odd numbers; the former relates to the Fourier analytic structure of the integers, but the latter relates to the Fourier analytic structure of the natural numbers, which is far worse.)