Volume growth of black hole interior

I suspect that Susskind arrived at the growth rate heuristically by considering the horizon as a codimension two hypersurface and then integrating over the radius of AdS space. Let's assume that the volume growth rate is constant with respect to Euclidean time to make the following argument. The SAdS$_4$ metric, and its volume form are given by $$text{d}s^2 =g^E_{munu}text{d} x^mutext{d} x^nu= f(r)text{d}t_{E}^2 + f(r)^{-1}text{dr}^2 + r^2 text{d}theta^2+r^2sin(theta)^2text{d}phi^2, sqrt{det{g}^E}varepsilon = r^2sin(theta)text{d}t_Ewedgetext{d}rwedgetext{d}thetawedgetext{d}phi.$$ Now consider the codimension two surface normal to the $t_E-r$ disk evaluated on the horizon given by $$text{d}s_partial^2 =h_{ab}text{d}x^atext{d}x^b= r_h^2text{d}theta^2 + r_h^2sin(theta)^2text{d}phi^2, sqrt{det{h}}varepsilon_partial = r_h^2sin(theta)text{d}thetawedgetext{d}phi.$$ If we integrate the codimension two volume form, we get the surface area of the SAdS$_4$ black hole, i.e., $$A_h = intsqrt{det{h}} varepsilon_partial = r_h^2int_0^pitext{d}thetaint_0^{2pi}text{d}phi sin(theta) = 4pi r_h^2.$$ Now, the trick is to consider a differential of the black hole volume as it grows in an interval $text{d}t_E$. We have $$dV = A_hdr implies A_h = frac{text{d}V}{text{d}r} = frac{text{d}V}{text{d}t_E}frac{text{d}t_E}{text{d}r} iff int A_h text{d}r = frac{text{d}V}{text{d}t_E}intfrac{text{d}t_E}{text{d}r}text{d}r = frac{text{d}V}{text{d}t_E}inttext{d}t_E $$ where I have used that the complexity growth rate is constant. Now, integrating over the Euclidean periodicity and over AdS space, I have $$A_h ell_{AdS} = frac{text{d}V}{text{d}t_E}int_0^betatext{d}t_E = beta frac{text{d}V}{text{d}t_E} implies frac{text{d}V}{text{d}t_E} = beta^{-1}A_hell_{AdS} = TA_hell_{AdS}.$$ That's my best bet anyway, you may have to dig deeper into literature to find out if those assumptions are the one's Susskind used but this at least gives you a start.