First note that because complex roots of a polynomial with real coefficients come in conjugate pairs, the roots of $P(z)$ are $\overline{z_1}, \overline{z_2}, \overline{z_3}, \overline{z_4}$. In other words, $\{z_1, z_2, z_3, z_4\} = \{\overline{z_1}, \overline{z_2}, \overline{z_3}, \overline{z_4}\}$. Thus $Q(z)$ is the polynomial
$$(z - 4iz_1)(z - 4iz_2)(z - 4iz_3)(z - 4iz_4).$$
It follows from Vieta's formulas that
$$B = (4iz_1)(4iz_2) + (4iz_1)(4iz_3) + (4iz_1)(4iz_4) + (4iz_2)(4iz_3) + (4iz_2)(4iz_4) + (4iz_3)(4iz_4)$$
and
$$D = (4iz_1)(4iz_2)(4iz_3)(4iz_4).$$

Applying Vieta's formulas to $P(z)$ yields
$$3 = z_1z_2 + z_1z_3 + z_1z_4 + z_2z_3 + z_2z_4 + z_3z_4 \quad \text{and} \quad 1 = z_1z_2z_3z_4.$$
Thus $B = (4i)^2 \cdot 3 = -16 \cdot 3 = -48$ and $D = (4i)^4 \cdot 1 = 256$. The requested sum is $B+D = -48+256 = 208$.

As above, $\{z_1, z_2, z_3, z_4\} = \{\overline{z_1}, \overline{z_2}, \overline{z_3}, \overline{z_4}\}$. Let $R(z) = (4i)^4 \cdot P(\frac{z}{4i})$. Then the roots of $R(z)$ are $f(z_j)$ for $j = 1, 2, 3, 4$ and its leading coefficient is 1, so $R(z) = Q(z)$. Therefore
$$\begin{aligned} A &= 4 \cdot (4i) = 16i \\ B &= 3 \cdot (4i)^2 = -48 \\ C &= 2 \cdot (4i)^3 = -128i \\ D &= (4i)^4 = 256. \end{aligned}$$
The requested sum is $-48 + 256 = 208$.