Two dimensional equable shapes are those shapes whose area and perimeter are numerically equal. There are only two rectangles that have the numerical equivalent of area and perimeter with positive integer lengths, the $3\times 6$ rectangle and the $4\times 4$ rectangle.
Proof: Let $d_1$ be the length of a rectangle on one dimension, $d_2$ be the length of a rectangle across the second dimension. WLOG let $d_1\ge d_2$. Then $2d_1+2d_2=d_1d_2 \Rightarrow \frac{1}{d_1}+\frac{1}{d_2}=\frac{1}{2}$, $\quad\frac{1}{2}>\frac{1}{d_2}\ge\frac{1}{4}\Rightarrow 2<d_2\le 4\Rightarrow d_2=3$ or $d_2=4$. Pluging these into the equation the original $3\times 6$ and $4\times 4$ rectangles are obtained.
For the three dimensional case there are ten cuboids whose surface area and volume are numerically equal.
Proof: Let $d_1$ be the length of a cubiod on one dimension, $d_2$ be the length of a cubiod across the second dimension, and $d_3$ be the length of a cubiod across the third dimension. WLOG let $d_1\ge d_2\ge d_3$. Then $2d_1d_2+2d_2d_3+2d_1d_3=d_1d_2d_3 \Rightarrow \frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}=\frac{1}{2}$, $\quad\frac{1}{2}>\frac{1}{d_3}\ge\frac{1}{6}\Rightarrow 2<d_2\le 6\Rightarrow d_3=3$ or $d_3=4$ or $d_3=5$ or $d_3=6$.
Case 1: ($d_3=3$) Replacing $d_3$ in the equation $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}=\frac{1}{2}$ gives the result $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{3}=\frac{1}{2}\Rightarrow\frac{1}{d_1}+\frac{1}{d_2}=\frac{1}{6}$, $\quad\frac{1}{6}>\frac{1}{d_2}\ge\frac{1}{12}\Rightarrow 6<d_2\le 12\Rightarrow d_2=7$ or $d_2=8$ or $d_2=9$ or $d_2=10$ or $d_2=11$ or $d_2=12$. $\quad \frac{1}{6}-\frac{1}{d_2}$ produces a unit fraction in all cases with the exception of $d_2=11$ which produces $\frac{5}{66}$. This results in five solutions (3,7,42),(3,8,24),(3,9,18),(3,10,15),(3,12,12)
Case 2: ($d_3=4$) Replacing $d_3$ in the equation $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}=\frac{1}{2}$ gives the result $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{4}=\frac{1}{2}\Rightarrow\frac{1}{d_1}+\frac{1}{d_2}=\frac{1}{4}$, $\quad\frac{1}{4}>\frac{1}{d_2}\ge\frac{1}{8}\Rightarrow 4<d_2\le 8\Rightarrow d_2=5$ or $d_2=6$ or $d_2=7$ or $d_2=8$. $\quad \frac{1}{6}-\frac{1}{d_2}$ produces a unit fraction in all cases with the exception of $d_2=7$ which produces $\frac{3}{28}$. This results in three solutions (4,5,20),(4,6,12),(4,8,8)
Case 3: ($d_3=5$) Replacing $d_3$ in the equation $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}=\frac{1}{2}$ gives the result $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{5}=\frac{1}{2}\Rightarrow\frac{1}{d_1}+\frac{1}{d_2}=\frac{3}{10}$, $\quad\frac{1}{5}\ge\frac{1}{d_2}>\frac{3}{20}\Rightarrow 5\ge d_2> \frac{20}{3}\Rightarrow d_2=5$ or $d_2=6$. $\quad \frac{3}{10}-\frac{1}{d_2}$ produces a unit fraction in all cases with the exception of $d_2=6$ which produces $\frac{2}{15}$. This results in one solution (5,5,10)
Case 4:($d_3=6$) Replacing $d_3$ in the equation $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}=\frac{1}{2}$ gives the result $\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{6}=\frac{1}{2}\Rightarrow\frac{1}{d_1}+\frac{1}{d_2}=\frac{1}{3}$, $\quad\frac{1}{6}\ge\frac{1}{d_2}\ge\frac{1}{6}\Rightarrow 6\ge d_2\ge 6\Rightarrow d_2=6$. This results in one solution (6,6,6)
I can continue this process to find the number of solutions for higher dimensions, however it gets very tedious. I found a lower bound for the number of n-dimensional hyper-rectangles with positive integer lengths, whose (n-1) dimensional surface area and n-dimensional volume is numerically equivalent, based off of the solutions of the previous dimension.
The smallest fraction in the solution of the previous dimension can be broken into two parts in several ways to produce new solutions in the next dimension. Let $L_k$ be the largest number in the $k^{th}$ solution of (n-1) dimensions. Let $f\in\Bbb{N}|\quad f|L_k$. We have $\frac{1}{L_k}=\frac{1}{L_k+f}+\frac{f}{(L_k+f)L_k}$. (Since $f|L_k$ the last fraction in the equation is a unreduced unit fraction) There are two more solutions in N dimensions which are (2N,2N,2N,…,2N[N times]) and (2N-1,2N-1,2N-1,…,2N-1[N-1 times],4N-2). We have the following lower bound formula:
$$S_N\ge 2+\sum_{k=1}^{S_{N-1}}d(L_k)$$
where $d(x)$ is the number of factors of $x$ and $S_N$ is the number of hyper-rectangles in n dimensions.
For example $S_3\ge 2+d(6)+d(4)=2+4+3=9$ ((3,10,15) isn’t counted)
Second example $S_4\ge 2+d(42)+d(24)+d(18)+d(15)+d(12)+d(20)+d(12)+d(8)+d(10)+d(6)=2+8+8+6+4+6+6+6+4+4+4=58$
question: can we find a formula for the number n-dimensional hyper-rectangles with positive integer lengths, whose (n-1) dimensional surface area and n-dimensional volume is numerically equivalent?
This is an equivalent question: can we find a formula for the number of ways $\frac{1}{2}$ can be written as the sum of n unit fractions?
Short answer: nobody knows.
In somewhat greater detail, the results for $2\le n\le6$ are tabulated at the Online Encyclopedia of Integer Sequences. The numbers are $2, 10, 108, 2892, 270332$. That's all anyone has been able to calculate. A reference is given to Gerald E. Gannon, Martin V. Bonsangue and Terrence J. Redfern, One Good Problem Leads to Another and Another and..., Math. Teacher, 90 (#3, 1997), pp. 188-191. The 108 solutions for $n=4$ are listed here.