Aren't you amazed at this?

Denote our Integral by I, then by noting the cyclic symmetry we can break it down into intervals as,

$\displaystyle \begin{aligned} &\int_0^1 \int_0^1 \int_0^1\left(\left\{\dfrac{x}{y}\right\}\left\{\dfrac{y}{z}\right\}\left\{\dfrac{z}{x}\right\}\right)^2 \; dx\; dy\; dz \\ &= 3\int\limits_{0<x<y<z<1}\left(\dfrac{x}{y}\dfrac{y}{z}\left\{\dfrac{z}{x}\right\}\right)^2 \; dx\; dy\; dz + 3\int\limits_{0<x<z<y<1} \left(\dfrac{x}{y}\left\{\dfrac{y}{z}\right\}\left\{\dfrac{z}{x}\right\}\right)^2 \; dx\; dy\; dz \\ &=I_1+I_2 \end{aligned}$

We will calculate $I_1$ and $I_2$ separately here,

$\displaystyle \begin{aligned} &I_1 \\ &= 3\int\limits_{0<x<y<z<1}\left(\dfrac{x}{y}\dfrac{y}{z}\left\{\dfrac{z}{x}\right\}\right)^2 \; dx\; dy\; dz \\ &= 3\int_{z=0}^1 \int_{y=x}^{z}\int_{x=0}^z \dfrac{x^2}{z^2} \left\{\dfrac{z}{x}\right\}^2 \; dx\; dy\; dz \\ &= 3 \int_{z=0}^1 \int_{x=0}^z \dfrac{(z-x)x^2}{z^2} \left\{\dfrac{z}{x}\right\}^2 \; dx\; dy\; dz \\ &= 3\int_{z=0}^1 \int_{t=0}^1 z^2 t^2(1-t) \left\{\dfrac{1}{t}\right\}^2 \; dt\; dz \quad \color{#3D99F6}{\text{By x=zt}} \\ &= \int_{z=0}^1 3z^2 \; dz \int_{t=0}^1 (t^2-t^3) \left\{\dfrac{1}{t}\right\}^2 \; dt \\ &= \int_{0}^1 t^2 \left\{\dfrac{1}{t}\right\}^2\; dt-\int_{0}^1 t^3 \left\{\dfrac{1}{t}\right\}^2\; dt \\ &= \mathfrak{S}_2-\mathfrak{S}_3 \end{aligned}$

Next for $I_2$ we have,

$\displaystyle \begin{aligned} &I_2 \\ &= 3\int\limits_{0<x<z<y<1} \left(\dfrac{x}{y}\left\{\dfrac{y}{z}\right\}\left\{\dfrac{z}{x}\right\}\right)^2 \; dx\; dy\; dz \\ &= 3\int_{y=0}^1 \int_{x=0}^z \int_{z=0}^y \left(\dfrac{x}{y}\left\{\dfrac{y}{z}\right\}\left\{\dfrac{z}{x}\right\}\right)^2 \; dx\; dy\; dz \\ &= 3\int_{y=0}^1 \int_{z=0}^y \left\{\dfrac{y}{z}\right\}^2 \dfrac{1}{y^2} \int_{x=0}^z x^2 \left\{\dfrac{z}{x}\right\}^2 \; dx\; dy\; dz \\ &= 3\int_{y=0}^1 \int_{z=0}^y \left\{\dfrac{y}{z}\right\}^2 \dfrac{z^3}{y^2} \int_{t=0}^1 t^2 \left\{\dfrac{1}{t}\right\}^2 \; dt \; dy\; dz \quad \color{#3D99F6}{\text{By x=zt}} \\ &= \mathfrak{S}_2 \int_{y=0}^1 3y^2 \; dy \int_{t=0}^1 t^3 \left\{\dfrac{1}{t}\right\}^2\; dt \quad \color{#3D99F6}{\text{By z=yt}} \\ &= \mathfrak{S_2}\mathfrak{S_3}\end{aligned}$

It remains to calculate $\mathfrak{S_2}$ and $\mathfrak{S_3}$ which can be evaluated as,

$\displaystyle \begin{aligned} &\mathfrak{S}_2 \\ &= \int_{t=0}^1 t^2 \left\{\dfrac{1}{t}\right\}^2\; dt \\ &= \int_{1}^\infty \dfrac{\{x\}^2}{x^4}\; dx \\ &= \sum_{k=1}^\infty \int_0^1 \dfrac{x^2}{(k+x)^4}\; dx \\ &= \sum_{k=1}^\infty \dfrac{1}{3k(k+1)^3} \\ &= 1-\dfrac{\zeta(2)}{3}-\dfrac{\zeta(3)}{3}\end{aligned}$

$\displaystyle \begin{aligned} &\mathfrak{S}_3 \\ &= \int_{t=0}^1 t^3 \left\{\dfrac{1}{t}\right\}^2\; dt \\ &= \int_{1}^\infty \dfrac{\{x\}^2}{x^5}\; dx \\ &= \sum_{k=1}^{\infty}\int_0^1 \dfrac{x^2}{(k+x)^5}\; dx \\ &= \sum_{k=1}^\infty \dfrac{4k+1}{12k^2(k+1)^4} \\ &= \dfrac{1}{2}-\dfrac{\zeta(3)}{6}-\dfrac{\zeta(4)}{4}\end{aligned}$

Now combining we have,

$\displaystyle \begin{aligned} &\int_0^1 \int_0^1 \int_0^1\left(\left\{\dfrac{x}{y}\right\}\left\{\dfrac{y}{z}\right\}\left\{\dfrac{z}{x}\right\}\right)^2 \; dx\; dy\; dz \\ &= I_1+I_2 \\ &= \mathfrak{S}_2-\mathfrak{S}_3+\mathfrak{S}_2\mathfrak{S}_3 \\ &= 1-\dfrac{\pi^2}{12}-\dfrac{\zeta(3)}{2}+\dfrac{\pi^2\zeta(3)}{108}+\dfrac{\pi^4 \zeta(3)}{1080}+\dfrac{\pi^6}{6480}+\dfrac{\zeta^2 (3)}{18}\end{aligned}$

This makes the answer $\boxed{1+12+3+2+3+108+6480+3+18+3+1080=7713}$