1. Let $P(x)=4x^4+8x^3-17x^2-12x+9$ with roots $x_1,x_2,x_3,x_4$.

We need to find $\prod_{i=1}^4 (4+x_i^2).$

2. Observe that $4+x_i^2=(x_i-2i)(x_i+2i).$ Hence, $\prod_{i=1}^4 (4+x_i^2)=\prod_{i=1}^4 (x_i-2i)\prod_{i=1}^4 (x_i+2i).$

3. Use the standard identity for a polynomial $P(x)=a\prod_{i=1}^4 (x-x_i),$ so that $P(2i)=a\prod_{i=1}^4 (2i-x_i), \qquad P(-2i)=a\prod_{i=1}^4 (-2i-x_i).$ Since $a=4$, $\prod_{i=1}^4 (x_i-2i)=\prod_{i=1}^4 (-(2i-x_i))=P(2i)/4$ because the degree is $4$ (even), and similarly, $\prod_{i=1}^4 (x_i+2i)=P(-2i)/4.$ Therefore, $\prod_{i=1}^4 (4+x_i^2)=\frac{P(2i)P(-2i)}{16}.$

4. Now compute $P(2i)$: $P(2i)=4(2i)^4+8(2i)^3-17(2i)^2-12(2i)+9.$ Using $(2i)^2=-4,\quad (2i)^3=-8i,\quad (2i)^4=16,$ we get $P(2i)=4(16)+8(-8i)-17(-4)-24i+9$ $=64-64i+68-24i+9$ $=141-88i.$ Similarly, $P(-2i)=141+88i.$

5. Hence, $P(2i)P(-2i)=(141-88i)(141+88i)=141^2+88^2.$ Compute: $141^2=19881, \qquad 88^2=7744,$ so $P(2i)P(-2i)=19881+7744=27625.$ Therefore, $\prod_{i=1}^4 (4+x_i^2)=\frac{27625}{16}.$

6. Given that $\left(4+x_1^2\right)\left(4+x_2^2\right)\left(4+x_3^2\right)\left(4+x_4^2\right)=\frac{125}{16}m,$ we compare: $\frac{125}{16}m=\frac{27625}{16}.$ Thus, $125m=27625$ $m=\frac{27625}{125}=221.$

7. Final answer: $\boxed{221}$

The derived answer matches the stored correct answer.