My question is what do we know about the product $\prod_{n\in \mathbb{N}}(1\frac{x^m}{n^m})?$ which is a slightly modified product from the eulerian product.

1$\begingroup$ For $m=2$, it's a wellknown expression for sinc; Mathematica will give you similar generalizations for fixed even $m$, and reciprocal $\Gamma$ functions for fixed odd $m$ ... not sure about a nice general expression for variable $m$ ... $\endgroup$ – Michael Engelhardt Aug 4 '19 at 18:36

1$\begingroup$ Also Euler product is for products over primes, Euler's formula for $\sin(x)$ is called Weierstrass product $\endgroup$ – reuns Aug 5 '19 at 3:43
Well, for $m=2$, we have the standard product $\prod_{n\in \mathbb N} (1\frac {x^2} {n^2})=\frac {\sin(\pi x)} {\pi x}$.
If we denote your product by $P_m(x)$, then we can evaluate eg. $P_4(x)$ like this:
$$(1\frac {x^4} {n^4}) = (1\frac {x^2} {n^2})(1+\frac {x^2} {n^2})=(1\frac {x^2} {n^2})(1\frac {(ix)^2} {n^2}).$$ Thus, $$P_4(x)=\frac {\sin(\pi x)} {\pi x} \frac {\sin(i \pi x)} {i\pi x}=\frac {\sin(\pi x) \sinh(\pi x)} {\pi^2 x^2}$$
We can write an interesting identity:
$$\log P_m(x)=\sum_{n\in\mathbb N} \log (1\frac {x^m} {n^m})=\sum_{n=1}^\infty\sum_{k=1}^\infty \frac 1 k (\frac x m)^{mk}=\sum_{k=1}^\infty \frac 1 k \zeta (mk) x^{mk}$$
By differentiating, we get
$$\frac {P'_m(x)} {P_m(x)}=\frac mx\sum_{k=1}^\infty \zeta(mk)x^{mk}$$
EDIT: Also, by using $1/\Gamma(z)=ze^{\gamma z}\Pi_{n\in \mathbb N} (1+\frac zn)e^{z/n}$ and expanding $(1z^3/n^3)$ as $(1z/n)(\omegaz/n)(\omega^2z/n)$, we get:
$$P_3(z)=\frac {1/z^3} {\Gamma(z)\Gamma(\omega z)\Gamma(\omega^2 z)}$$
This is similar to
$$P_2(z) = \frac {1/z^2} {\Gamma(z)\Gamma(z)} = \frac {\sin(\pi z)} {\pi z}$$ (by the reflection formula)
I think it is easy to see now that those formulas generalize to all $m$. We just take $1/z^m$ (or $1/z^m$ for $m$ even) and divide by the product of $\Gamma(\rho z)$, where $\rho$ goes over all $m$th roots of unity.
$$P_m(z)=(1)^{m+1}\frac 1 {z^m} \prod_\rho \frac 1 {\Gamma(\rho z)}$$
For even $m$, this can be further reduced using the reflection formula, because the roots of unity form pairs (both $\rho$ and $\rho$ are roots of unity).
The even case $$\prod_{n \ge 1} (1(x/n)^{2m}) =\prod_{a=1}^m \prod_{n \ge 1} (1 e^{2i \pi a/m}(x/n)^2)= \prod_{a=1}^m \frac{\sin(\pi e^{i \pi a/ m} x)}{\pi e^{i \pi a/ m} x}$$
Otherwise $$\prod_{n \ge 1} (1(x/n)^m) =\lim_{N \to \infty }\prod_{a=1}^m \prod_{n=1}^N (1 e^{2i \pi a/m}x/n)= \prod_{a=1}^m \frac{e^{2i \pi a/ m} x}{\Gamma(e^{2i \pi a/ m} x)}$$