Imagine I have a pointlike Brownian particle, with diffusion constant $D$, and I place it at some initial coordinate in a cage of known geometry. Assuming the volume $V$ of the cage is "everywhere" accessible, how long does it take the particle to lose its "memory" of the initialization coordinate and achieve a uniform probability distribution across the cage? How precisely can one measure this, and how important is the geometry of the cage?

Let $\rho(t,x)$ be the probability that the particle is at location $x$ at time $t$. $\rho$ satisfies the equation $\rho_t = \rho_{xx}$, with $\rho(0,x)=\delta(xx_0)$ where $x_0$ is the starting position. You have Neumann boundary conditions on the boundary of your domain. Suppose the eigenfunctions of the negative Laplacian of your domain with Neumann boundary conditions are $\phi_0, \phi_1, \ldots$ with corresponding eigenvalues $\lambda_0 \leq \lambda_1 \leq \lambda_2 \cdots$. $\phi_1$ is constant with $\lambda_0=0$. The solution to the equation is $$ \rho(t,x)= \sum_{j=0}^\infty c_j e^{ \lambda_j t} \phi_j (x) $$ where $c_j$ is the inner product of $\delta(xx_0)$ with $\phi_j$ which gives $c_j=\phi_j(x_0)$. When $t$ goes to infinity you just get a constant for $\rho$. Your question is (I think) equivalent to asking how long does it take for all the $j>0$ terms to die out. Assuming $c_1 \neq 0$, this will be determined by $\lambda_1$; a bigger $\lambda_1$ means faster approach to uniform probability. There is not going to be an explicit formula for $\lambda_1$ for most domains. But you can get some intuition. As Alice says, if you have two equal volumes with a narrow connection, then $\lambda_1$ is going to be very small and it will take a long time for $\rho$ to be flat. (Unless you start exactly midway between the two volumes. This corresponds to $c_1=0$.) 


Seems like the timescale should depend on the number and sizes of spherical balls it takes to cover the entire space $V$. For example if your space has a narrow neck connecting two balloons then it would take a while for your particle to diffuse from the balloon where it started through the neck to the other side. I would have entered as a comment but can't yet (sorry). 


For the initial distribution being delta function $\delta(y)$ the result will be the Green function of the Neumann problem for the heat equation. If the domain is regular enough the function can be written as $$ G(x,y,t)=\sum_{n=1}^\infty \varphi_n(x)\varphi_n(y)e^{\lambda_n t} $$ where $\lambda_n$ and $\varphi_n$ are eigenvalues and normed in $L_2$ eigenfunctions of the corresponding elliptic Neumann problem. The first eigenvalue $\lambda_1=0$. So the rate of convergence to constant is exponential and determined by the next value $\lambda_2>0$. 

