Let $a,b,c,d,e,f$ be real numbers such that the polynomial
\[ p(x)=x^8-4x^7+7x^6+ax^5+bx^4+cx^3+dx^2+ex+f \]
factorises into eight linear factors $x-x_i$, with $x_i>0$ for $i=1,2,\ldots,8$. Determine all possible values of $f$.
1. **Identify the polynomial and its roots:**
   Given the polynomial:
   \[
   p(x) = x^8 - 4x^7 + 7x^6 + ax^5 + bx^4 + cx^3 + dx^2 + ex + f
   \]
   which factorizes into eight linear factors \(x - x_i\) with \(x_i > 0\) for \(i = 1, 2, \ldots, 8\). Therefore, we can write:
   \[
   p(x) = (x - x_1)(x - x_2) \cdots (x - x_8)
   \]

2. **Use Newton's Identities:**
   Let \(S_n = \sum_{i=1}^8 x_i^n\), where \(x_i\) are the roots of \(p(x)\). By Newton's identities, we have:
   \[
   S_1 - 4 = 0 \implies S_1 = 4
   \]
   \[
   S_2 - 4S_1 + 14 = 0 \implies S_2 - 4 \cdot 4 + 14 = 0 \implies S_2 = 2
   \]

3. **Apply the Quadratic Mean - Arithmetic Mean (QM-AM) Inequality:**
   From the QM-AM inequality, we have:
   \[
   \sqrt{\frac{1}{8} \sum_{i=1}^8 x_i^2} \geq \frac{1}{8} \sum_{i=1}^8 x_i
   \]
   Substituting \(S_1\) and \(S_2\):
   \[
   \sqrt{\frac{1}{8} \cdot 2} \geq \frac{4}{8}
   \]
   \[
   \sqrt{\frac{1}{4}} \geq \frac{1}{2}
   \]
   \[
   \frac{1}{2} \geq \frac{1}{2}
   \]
   This equality holds if and only if all \(x_i\) are equal. Therefore, \(x_1 = x_2 = \cdots = x_8\).

4. **Determine the value of the roots:**
   Since \(S_1 = 4\) and all roots are equal, let \(x_i = r\) for all \(i\). Then:
   \[
   8r = 4 \implies r = \frac{1}{2}
   \]

5. **Calculate the value of \(f\):**
   The constant term \(f\) of the polynomial is the product of all roots:
   \[
   f = \prod_{i=1}^8 x_i = \left(\frac{1}{2}\right)^8 = \frac{1}{256}
   \]

The final answer is \(\boxed{\frac{1}{256}}\)