How can you estimate how much of a substance will dissolve in a given lipid?
The major issue with the blood-brain barrier is that aside from small hydrophobic/lipophilic compounds (including $\ce{O_{2}}$ and $\ce{CO_{2}}$), almost everything else is prevented from passing due to cellular tight junctions and will only (potentially) pass through via active transport.

However, for small lipophilic molecules passing through a phospholipid bilayer (such as on a cell surface in the blood-brain barrier), this post provides information on equations: http://www.ncbi.nlm.nih.gov/books/NBK21626/

Where:

$\frac{dn}{dt}$ is diffusion rate (mol/s)

$K$ is the partition coefficient (the equilibrium constant for its partition between oil and water) - e.g. for Urea, K = 0.0002 while for diethylurea which is more hydrophobic, K = 0.01

$C^{m}$ is the concentration in the hydrophobic part of the phosopholipid bilayer (mol/L)

$C^{aq}$ is the concentration in aqueous solution (mol/L)

$P$ is permeability coefficient

$D$ is diffusion coefficient of the substance within the phospholipid membrane

$A$ is the area of the membrane (in $m^{2}$ I think?)

$x$ is the membrane thickness (in m I think?)

$C_{1}^{aq}$ and $C_{2}^{aq}$ are the aqueous concentrations on either side of the layer (one will be the serum or CSF concentration, the other will be the intracellular concentration)


$K = \frac{C^{m}}{C^{aq}}$
$\frac{dn}{dt}=PA(C_{1}^{aq}-C_{2}^{aq})$ and $P = \frac{KD}{x}$
$\frac{dn}{dt}=A\frac{KD}{x}(C_{1}^{aq}-C_{2}^{aq})$




You would have a similar equation to 2. on the other side of the cells lining the blood-brain barrier.

It seems from my reading that $P$ tends to be better known than $D$, which is actually usually calculated using equation 2. from $P$, $K$ and $x$.

Hope that helps!