I know that the gradient descent flow of the Dirichlet energy $$\min_u E(u) = \int_{\Omega}|\nabla u|^2 dA$$ Is the diffusion/heat equation:
$$u_t = \Delta u$$
Is there a change in the Dirichlet energy such that it gives an anisotropic diffusion flow?
Sure. Just make the obvious change to the Dirichlet energy. Let $A\in\mathcal{C}^1(\Omega;\mathbb{R}^{n\times n})$ be a matrix-valued function that is symmetric and positive definite for every $x\in\Omega$ (I think the regularity of $A$ can be relaxed to only having measurable coefficients, but I'm not too familiar with the intricacies of regularity theory in this context).
Now consider $E:L^2(\Omega)\to\overline{\mathbb{R}}$ be defined as $$E_{A}(u):=\int_{\Omega}A\nabla u\cdot\nabla u~dx$$ for $u\in H^1(\Omega)$ and extended to $+\infty$ for all other $u\in L^2(\Omega)\setminus H^1(\Omega)$.
One easily computes the Fréchet differential (Euler-Lagrange equations) of $E$ as $$dE_A(u)[v]=\int_{\Omega}A\nabla u\cdot\nabla v~dx$$ for $u,v\in H^1(\Omega)$. Thus, $dE_A(u)[v]=(u,v)_{L^2(\Omega)}$ implies that $-\operatorname{div}(A\nabla u)=v$ in the weak sense (essentially by definition). Now we see that the $L^2$-gradient flow of $E_A$ is nothing more than the desired anisotropic diffusion: $$\partial_tu-\operatorname{div}(A\nabla u)=0.$$
Note that here we are considering an $L^2$-gradient flow. For full generality / to rigorously define and analyze this flow you should consider this as an $L^2$-subgradient inclusion/flow since the energy is infinite for $u\in L^2(\Omega)\setminus H^1(\Omega)$. You don't even need to start with $u_0\in H^1(\Omega)$ for solutions to exist at all...
Yes, multiply the gradient with some matrix. For example an outer product tensor $${\bf T} = \sum_{\forall n} \lambda_n {\bf \hat e}_n {{\bf \hat e}_n}^T$$ and then modify like this: $$\min_u E(u) = \int_{\Omega}|{\bf T}\nabla u|^2 dA$$
This will make cost of flow in directions to depend on the eigensystem of $\bf T$. Of course $\bf T$ can be a function of the dimensions of space. For example if you have "conducting" fibers in some direction but "insulation" in another, maybe those fibers bend according to some curve or trajectory as you move over the spatial dimensions.
