By Vieta's Formulas,
$$\begin{aligned} p + q + r &= -2, \\ pq + pr + qr &= -1, \text{ and} \\ pqr &= -3. \end{aligned}$$
The expression to be evaluated is
$(p^2 + 4)(q^2 + 4)(r^2 + 4) = p^2q^2r^2 + 4(p^2q^2 + p^2r^2 + q^2r^2) + 16(p^2 + q^2 + r^2) + 64.$
The first term is
$$p^2q^2r^2 = (pqr)^2 = (-3)^2 = 9.$$
To find the second term, note that
$(pq + pr + qr)^2 = p^2q^2 + p^2r^2 + q^2r^2 + 2pqr(p + q + r).$
Rearranging and substituting yields
$$p^2q^2 + p^2r^2 + q^2r^2 = (-1)^2 - 2(-3)(-2) = -11.$$
For the third term,
$(p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + pr + qr),$
so
$$p^2 + q^2 + r^2 = (-2)^2 - 2(-1) = 6.$$
Therefore
$(p^2 + 4)(q^2 + 4)(r^2 + 4) = 9 + 4 \cdot (-11) + 16 \cdot 6 + 64 = 125.$

With $i = \sqrt{-1}$, the given expression can be factored as
$$(p^2 + 4)(q^2 + 4)(r^2 + 4) = (p - 2i)(p + 2i)(q - 2i)(q + 2i)(r - 2i)(r + 2i) \\ = [(p - 2i)(q - 2i)(r - 2i)] \cdot [(p + 2i)(q + 2i)(r + 2i)].$$
Let $f(x)$ be the given polynomial. Then $f(x) = -(p-x)(q-x)(r-x)$ and
$$-f(2i) = (p - 2i)(q - 2i)(r - 2i),$$
which is the first bracketed expression above. Furthermore
$$f(-x) = (-x - p)(-x - q)(-x - r) = -(p + x)(q + x)(r + x).$$
Therefore
$$-f(-x) = (p + x)(q + x)(r + x).$$
It follows that
$$-f(-2i) = (p + 2i)(q + 2i)(r + 2i),$$
the second bracketed expression above. Hence
$$(p^2 + 4)(q^2 + 4)(r^2 + 4) = -f(2i) \cdot (-f(-2i)) = f(2i) \cdot f(-2i).$$
Because
$$f(2i) = (2i)^3 + 2(2i)^2 - (2i) + 3 = -8i - 8 - 2i + 3 = -10i - 5$$
and
$$f(-2i) = (-2i)^3 + 2(-2i)^2 - (-2i) + 3 = 8i - 8 + 2i + 3 = 10i - 5,$$
the requested product is
$$f(2i) \cdot f(-2i) = (-10i - 5)(10i - 5) = 125.$$