XlogP of octanol and water itself
These values are consistent with the slight solubility of water in octanol and octanol in water. 

For example, the solubility of 1-octanol in water is 0.46 g/L. Assuming the density of a saturated solution of octanol in water is approximately that of water, then the concentration in grams per liter can be converted to moles per liter:

$$\dfrac{0.46\ \mathrm{g\ octanol}}{\mathrm{L\ water}}\times \dfrac{\mathrm{mol\ octanol}}{130.23\ \mathrm{g\ octanol}}=3.53\times 10^{-3}\ \mathrm{M}$$

The molar concentration of pure 1-octanol can be calculated from its density:

$$\dfrac{824 \ \mathrm{g\ octanol}}{\mathrm{L\ octanol}}\times \dfrac{\mathrm{mol\ octanol}}{130.23\ \mathrm{g\ octanol}}=6.33\ \mathrm{M}$$

And the $\log P$ is:

$$\log P = \log \left( \dfrac{\mathrm{[octanol]_{octanol}}}{\mathrm{[octanol]_{water}}} \right) \log \left( \dfrac{6.33\ \mathrm{M}}{3.53\times 10^{-3}\ \mathrm{M}} \right)=3.25 $$

Likewise, if we knew the solubility of water in 1-octanol, we could do the same calculation.