The derivation of the Cramer-Rao lower bound in Kay uses the weighted Cauchy-Schwarz inequality: $$ \left[ \int w(\mathbf{x})g(\mathbf{x})h(\mathbf{x})d\mathbf{x} \right]^2 \leq \int w(\mathbf{x})g^2(\mathbf{x})d\mathbf{x} \int w(\mathbf{x}) h^2(\mathbf{x}) d\mathbf{x} $$
where $g$ and $h$ are arbitrary scalar functions, and $w(\mathbf{x}) \geq 0$ for all $\mathbf{x}$. I don't have a source, but it seems reasonable that the same inequality would hold for absolute values (may be a special case of the more general Holder inequality?):
$$ \left| \int w(\mathbf{x})g(\mathbf{x})h(\mathbf{x})d\mathbf{x} \right| \leq \int w(\mathbf{x})\left|g(\mathbf{x})\right|d\mathbf{x} \int w(\mathbf{x}) \left|h(\mathbf{x}) \right| d\mathbf{x} $$
If the estimator is unbiased: $$ E[\hat{\theta}] = \theta $$ or $$ \int \hat{\theta} \: p(\mathbf{x};\theta) \: d\mathbf{x} = \theta $$
Differentiating with respect to $\theta$ and using $\dfrac{\partial p(\mathbf{x}; \theta)}{\partial \theta} = \dfrac{\partial \ln p(\mathbf{x};\theta)}{\partial \theta} p(\mathbf{x};\theta) $ yields:
$$ \int \hat{\theta} \: \dfrac{\partial \ln p(\mathbf{x}; \theta)}{\partial \theta} p(\mathbf{x}; \theta) \: d\mathbf{x} = 1 $$
$\hat{\theta}$ in this expression can be replaced with $(\hat{\theta} - \theta)$ because the CRLB assumes the regularity condition $\displaystyle E\left[\frac{\partial \ln p(\mathbf{x}; \theta)}{\partial \theta}\right] = 0$, so $ \displaystyle \int \theta \: \dfrac{\partial \ln p(\mathbf{x}; \theta)}{\partial \theta} p(\mathbf{x}; \theta) \: d\mathbf{x} = \theta \: E\left[\dfrac{\partial \ln p(\mathbf{x}; \theta)}{\partial \theta}\right] = 0 $ and then we have:
$$ \int (\hat{\theta} - \theta) \: \dfrac{\partial \ln p(\mathbf{x}; \theta)}{\partial \theta} p(\mathbf{x}; \theta) \: d\mathbf{x} = 1 $$
Here our derivation differs from Kay by applying the modified Cauchy-Schwartz inequality with $w(\mathbf{x}) = p(\mathbf{x};\theta)$, $g(\mathbf{x}) = \hat{\theta} - \theta$, and $h(\mathbf{x}) = \dfrac{\partial \ln p(\mathbf{x};\theta}{\partial \theta}$
$$ 1 \leq \int \left|(\hat{\theta} - \theta)\right| p(\mathbf{x}; \theta) \: d\mathbf{x} \int \left| \frac{\partial \ln p(\mathbf{x}; \theta)}{\partial \theta} \right| p(\mathbf{x}; \theta) d\mathbf{x} $$
Rearranging: $$ E\left[ \left|(\hat{\theta} -\theta)\right| \right] \geq \frac{1}{E\left[\left|\dfrac{\partial \ln p(\mathbf{x}; \theta)}{\partial \theta}\right| \right]} $$
The expectation in the denominator is no longer the Fisher Information $I(\theta)$