Since the coefficients of $ f(x) $ are all real, the nonreal roots come in conjugate pairs. Without loss of generality, assume that $ a_1 + ib_1 $ and $ a_2 + ib_2 $ are conjugates, and that $ a_3 + ib_3 $ and $ a_4 + ib_4 $ are conjugates, so $ a_1 = a_2, $ $ b_1 = -b_2, $ $ a_3 = a_4, $ and $ b_3 = -b_4 $. Then by Vieta's formulas, the product of the roots is \begin{align*} (a_1 + ib_1)(a_2 + ib_2)(a_3 + ib_3)(a_4 + ib_4) &= (a_1 + ib_1)(a_1 - ib_1)(a_3 + ib_3)(a_3 - ib_3) \\ &= (a_1^2 + b_1^2)(a_3^2 + b_3^2) \\ &= 65.\end{align*}The only ways to write 65 as the product of two positive integers are $ 1 \times 65 $ and $ 5 \times 13 $. If one of the factors $ a_1^2 + b_1^2 $ or $ a_3^2 + b_3^2 $ is equal to 1, then $ f(x) $ must have a root of $ \pm i $. (Remember that none of the roots of $ f(x) $ are real.) We can check that $ \pm i $ cannot be roots, so 65 must split as $ 5 \times 13 $. Wihtout loss of generality, assume that $ a_1^2 + b_1^2 = 5 $ and $ a_3^2 + b_3^2 = 13 $. Hence, $ \{|a_1|,|b_1|\} = \{1,2\} $ and $ \{|a_3|,|b_3|\} = \{2,3\} $. By Vieta's formulas, the sum of the roots is \begin{align*} (a_1 + ib_1) + (a_2 + ib_2) + (a_3 + ib_3) + (a_4 + ib_4) &= (a_1 + ib_1) + (a_1 - ib_1) + (a_3 + ib_3) + (a_3 - ib_3) \\ &= 2a_1 + 2a_3 = 6, \end{align*}so $ a_1 + a_3 = 3 $. The only possibility is that $ a_1 = 1 $ and $ a_3 = 2 $. Then $ \{b_1,b_2\} = \{2,-2\} $ and $ \{b_3,b_4\} = \{3,-3\}, $ so the roots are $ 1 + 2i, $ $ 1 - 2i, $ $ 2 + 3i, $ and $ 2 - 3i $. Then \begin{align*} f(x) &= (x - 1 - 2i)(x - 1 + 2i)(x - 2 - 3i)(x - 2 + 3i) \\ &= [(x - 1)^2 + 4][(x - 2)^2 + 9] \\ &= x^4 - 6x^3 + 26x^2 - 46x + 65.\end{align*}Therefore, $ p = \boxed{-46} $.