Rumination: Skip to the solution if you do not want to read what led me to the solution. I see a polynomial whose eight roots are positive, and yet only three coefficients are known. These few conditions lead me to suspect that some inequality is at work, perhaps AM-GM or Cauchy Schwarz.
Solution:
The given polynomial has positive zeros $x_{i}$, where $i=1,2,3,\dots, 8.$ Using Vieta's we have
$$ \sum_{i=1}^{8} x_{i} = 4, \ \text{and} \tag{A} $$
$$ \sum_{1\le i < j \le 8} x_{i}x_{j} =7 \tag{B} $$
Using $(A)$ and $(B)$ we have the following.
$$ \sum_{i=1}^8 x_{i}^2 = \left(\sum_{i=1}^8 x_{i} \right)^2 -2 \sum_{1\le i < j \le 8} x_{i}x_{j} = (4)^2 - 2 \cdot 7 =2 \tag{C} $$
Using the Cauchy-Schwarz Inequality, we have
$$ (1^2+1^2+\dots+1^2)(x_{1}^2+x_{2}^2+\dots +x_{8}^2) \ge (x_{1}+x_{2}+\dots +x_{8})^2 \tag{D} $$
Using $(A),(B),(C)$ we get that equality holds in $(D).$ Therefore,
$$ x_{1}=x_{2} = \dots = x_{8} =\frac{1}{2}. $$
Using Vieta's again
$$ f = \prod_{i=1}^8 x_{i} = \left(\frac{1}{2} \right)^8 \qed $$
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