We shall prove that $f(n) = F_{2n+1}$, the $(2n+1)$th Fibonacci number. Let $\zeta = e^{\frac{2\pi i}{2n+1}}$ be a primitive $(2n+1)$th root of unity. Then
$$1 + 4 \cos^2 \left( \frac{k\pi}{2n+1} \right) = 3 + 2 \cos \left( \frac{2k\pi}{2n+1} \right) = 3 + 2 \operatorname{Re}(\zeta^k) = 3 + \zeta^k + \zeta^{-k}.$$

Note that
$$\begin{align*} x^{2n+1} - 1 &= \prod_{k=1}^{2n+1} (x - \zeta^k) \\ &= (x - 1) \prod_{k=1}^{n} (x - \zeta^k)(x - \zeta^{-k}) \\ &= (x - 1) \prod_{k=1}^{n} (x^2 - (\zeta^k + \zeta^{-k})x + 1). \end{align*}$$
Let $\alpha = \frac{3+\sqrt{5}}{2}$ so that $\alpha^2 - 3\alpha + 1 = 0$. Putting $x = -\alpha$, we obtain
$$-\alpha^{2n+1} - 1 = (-\alpha - 1) \prod_{k=1}^{n} (\alpha^2 + (\zeta^k + \zeta^{-k})\alpha + 1) = - (\alpha + 1) \alpha^n \prod_{k=1}^{n} (3 + \zeta^k + \zeta^{-k}).$$
It follows that
$$f(n) = \prod_{k=1}^{n} (3 + \zeta^k + \zeta^{-k}) = \frac{\alpha^{2n+1} + 1}{\alpha^n(\alpha + 1)}.$$
Next, the terms of the Fibonacci sequence satisfy
$$F_{2n+5} = F_{2n+4} + F_{2n+3} = 2F_{2n+3} + F_{2n+2} = 3F_{2n+3} - F_{2n+1}.$$
Letting $a_n = F_{2n+1}$, we have the recurrence relation $a_{n+2} = 3a_{n+1} - a_n$ and the conditions $a_0 = 1$ and $a_1 = 2$. Since the roots to $\lambda^2 - 3\lambda + 1 = 0$ are $\alpha$ and $\frac{1}{\alpha}$, the general term is $a_n = A\alpha^n + B \cdot \frac{1}{\alpha^n}$ for some constants $A$ and $B$. Using $a_0 = 1$ and $a_1 = 2$, it is easy to deduce
$$a_n = \frac{\alpha}{\alpha + 1} \cdot \alpha^n + \frac{1}{\alpha + 1} \cdot \frac{1}{\alpha^n} = \frac{\alpha^{2n+1} + 1}{\alpha^n(\alpha + 1)} = f(n).$$
Therefore, $f(n)$ is an integer for all positive integers $n$.