Let $u(r,t)$ solve the heat equation in polar coordinates in the complement of the disk: $$ \partial_t u(r,t)= \frac{1}{2}\frac{1}{r}\partial_r r \partial_r u(r,t), \quad r>1$$ with initial condition $u(r,0)=0$ and boundary condition $u(1,t)=1$. I claim that the number you seek is $$ A_t = \pi + 2\pi\int_1^\infty u(r,t)r dr.$$ (This is because $u(r,t)$ gives the probability that a Brownian particle released at radius $r$ has hit the disk by time $t$. The first term $\pi$ is just the area of the disk at time $0$. I have transferred the problem to one in which your particle remains fixed and potential target points move.)
There is an exact expression for $u$ (see Wendel, J. G. "Hitting spheres with Brownian motion". Ann. Prob. 8, 164 (1980)., which I learned of from the answer to this mathoverflow question, but ultimately traced back to a paper of Carslaw and Jaeger from 1940). It is most easily expressed in terms of the Laplace transform $F(r,\lambda)=\int_0^\infty u(r,t) e^{-\lambda t}$: $$F(r,\lambda) = \frac{K_0(\sqrt{2\lambda}r)}{\lambda K_0(\sqrt{2\lambda})}, \quad r\ge 1,$$ where $K_0$ is a modified Bessel function of the second kind and order $0$. In your case the integral over $r$ can be done exactly since $\int r K_0(r) dr = - K_1(r)$. So $$ \int_0^\infty A_t e^{-\lambda t} d t \ = \ \frac{\pi}{\lambda} + \frac{2\pi K_1(\sqrt{2\lambda})}{\lambda (2\lambda)^{\frac{3}{2}}K_0(\sqrt{2\lambda})}.$$ You can recover $A_t$ by analytic continuation of the right hand side to complex $\lambda$ and Fourier inversion.