From
$$x^{8} - 4x^{7} + 7x^{6} + a x^{5} + b x^{4} + c x^{3} + d x^{2} + e x + f = (x - x_{1})(x - x_{2}) \ldots (x - x_{8})$$
we have
$$\sum_{i=1}^{8} x_{i} = 4 \quad \text{and} \quad \sum x_{i} x_{j} = 7$$
where the second sum is over all pairs $(i, j)$ of integers where $1 \leq i < j \leq 8$. Since this sum can also be written
$$\frac{1}{2}\left[\left(\sum_{i=1}^{8} x_{i}\right)^{2} - \sum_{i=1}^{8} x_{i}^{2}\right]$$
we get
$$14 = \left(\sum_{i=1}^{8} x_{i}\right)^{2} - \sum_{i=1}^{8} x_{i}^{2} = 16 - \sum_{i=1}^{8} x_{i}^{2}$$
so
$$\begin{equation*} \sum_{i=1}^{8} x_{i}^{2} = 2 \quad \text{while} \quad \sum_{i=1}^{8} x_{i} = 4 \quad [3 \text{ marks }] \tag{1} \end{equation*}$$
Now
$$\sum_{i=1}^{8} (2x_{i} - 1)^{2} = 4 \sum_{i=1}^{8} x_{i}^{2} - 4 \sum_{i=1}^{8} x_{i} + 8 = 4(2) - 4(4) + 8 = 0$$
which forces $x_{i} = 1/2$ for all $i$. [3 marks] Therefore
$$f = \prod_{i=1}^{8} x_{i} = \left(\frac{1}{2}\right)^{8} = \frac{1}{256}. \quad [1 \text{ mark}]$$

Alternate solution: After obtaining (1) [3 marks], use Cauchy's inequality to get
$$16 = (x_{1} \cdot 1 + x_{2} \cdot 1 + \cdots + x_{8} \cdot 1)^{2} \leq (x_{1}^{2} + x_{2}^{2} + \cdots + x_{8}^{2})(1^{2} + 1^{2} + \cdots + 1^{2}) = 8 \cdot 2 = 16;$$
or the power mean inequality to get
$$\frac{1}{2} = \frac{1}{8} \sum_{i=1}^{8} x_{i} \leq \left(\frac{1}{8} \sum_{i=1}^{8} x_{i}^{2}\right)^{1/2} = \frac{1}{2}. \quad [2 \text{ marks}]$$
Either way, equality must hold, which can only happen if all the terms $x_{i}$ are equal, that is, if $x_{i} = 1/2$ for all $i$. [1 mark] Thus $f = 1/256$ as above. [1 mark]