From: https://www.youtube.com/watch?v=DgmuGqeRTto

Consider an \(n\)-dimensional ball of fluid with size scale \(L\). The total kinetic energy within such a ball is of the order of $$ v^2 L^n. $$ Assuming the evolution of the fluid ball consume energy, we have $$ v \propto L^{-n/2}. $$ Thus the Reynolds number defined as $$ \text{Re} = \frac{v L} {\nu} $$ will be $$ \text{Re} \propto L^{1-n/2}. $$ Hence, for \(n=2\), \(\text{Re}\) does not depend on the length scale, which means that turbulence will not be amplified at small scale. For \(n=3\), \(\text{Re}\propto L^{-1/2}\), which means that turbulence becomes stronger at small scale.