Let $$f_n(x) = \left| \prod_{k=0}^n \left( x - \frac{k}{n} \right) \right|^{\frac{1}{n}}$$ Based purely on examination of the graph below, which has $n=100$, it appears that for $x \not \in \mathbb{Q} \cap [0, 1], \; \, f_n(x) \to f(x)$ as $n \to \infty$ where $f$ is some continuous function.
I have been attempting to work towards identifying $f$, although unfortunately haven't really gotten anywhere. Otherwise some insight into proving (or disproving) any of the following observations would also be useful:
I have also attempted to consider $\log f_n(x) = \frac{1}{n} \sum_{k=0}^n \log |x-\frac{k}{n}|$, but to no avail.
Let $f$ be the function represented by
$$f(x)=\lim_{n\to \infty}\left|\prod_{k=0}^n\left(x-\frac kn\right)\right|^{1/n}$$
Next, note that
$$\left|\prod_{k=0}^n\left(x-\frac kn\right)\right|^{1/n}=e^{\frac1n \sum_{k=0}^n \log\left|\left(x-\frac kn\right)\right|}$$
The limit of the exponent is the Riemann sum for $\int_0^1 \log|x-t|\,dt$, when $x\in [0,1]$ and is irrational. For $x\in [0,1]$ and irrational, we have
$$\begin{align} \int_0^1 \log|x-t|\,dt&=\lim_{\epsilon\to 0^+}\left(\int_0^{x-\epsilon} \log(x-t)\,dt+\int_{x+\epsilon}^1 \log(t-x)\,dt\right)\\\\ &=x\log(x)-x+(1-x)\log(1-x)-(1-x)\\\\ &=x\log(x)+(1-x)\log(1-x)-1 \end{align}$$
and $f(x)=e^{-1}x^x\,(1-x)^{1-x}$.
If $x>1$, then $f(x)=e^{\int_0^1 \log(x-t)\,dt}=e^{-1}\frac{x^x}{(x-1)^{x-1}}$.
If $x<0$, then $f(x)=e^{\int_0^1 \log(t-x)\,dt}=e^{-1}\frac{(1-x)^{1-x}}{(-x)^{-x}}$.