Determinant of the "real part" of a matrix

Bound 1 does not hold already for $n=2$. Take a matrix $A=\pmatrix{e^{ia}&e^{ib}\\e^{ic}&e^{id}}$, it satisfies your conditions if $|a+d-b-c|\leqslant \pi/3$. On the other hand, $X=\pmatrix{\cos a&\cos b\\\cos c&\cos d}$ and $$\det X=\cos a\cos d-\cos b\cos c=\frac12\left(\cos(a+d)+\cos(a-d)-\cos(b+c)-\cos(b-c)\right)\\= \frac12\left(2\sin\frac{b+c-a-d}2\sin\frac{a+d+b+c}2+\cos(a-d)-\cos(b-c)\right), $$ thus if $a-d=0$, $b-c=\pi$, $a+d+b+c=\pi$, $b+c-a-d=\pi/3$ (I am lazy to solve this explicitly), this expression is equal to $3/2>1$.

As a complement to Fedor's answer, let me note the following special case, where the desired claim does hold.

Prop. Let $A$ be a complex matrix and $A=X+iY$ be its Cartesian decomposition, i.e., $X=\frac12(A+A^*)$ and $Y=\frac{1}{2i}(A-A^*)$. If $X>0$ (i.e., it is positive definite), then in fact $|\det A| \ge \det X$.