1. We are given $f(x)=x^5+2e^{x/4}$ and a function $g$ such that $(g\circ f)(x)=g(f(x))=x\quad \text{for all }x\in\mathbb R.$ So $g$ is the inverse function of $f$.

2. Differentiate the identity $g(f(x))=x$ using the chain rule: $g'(f(x))\,f'(x)=1.$ Hence, $g'(f(x))=\frac{1}{f'(x)}.$

3. We need $g'(2)$. So first find the value of $x$ such that $f(x)=2.$ Solve: $x^5+2e^{x/4}=2.$ Try $x=0$: $f(0)=0^5+2e^0=2.$ So $f(0)=2$. Therefore, $g'(2)=g'(f(0))=\frac{1}{f'(0)}.$

4. Compute $f'(x)$: $f'(x)=5x^4+2\cdot \frac14 e^{x/4}=5x^4+\frac12 e^{x/4}.$ Thus, $f'(0)=5\cdot 0^4+\frac12 e^0=\frac12.$ So, $g'(2)=\frac{1}{1/2}=2.$

5. Therefore, $8g'(2)=8\times 2=16.$

So the correct option is C.