This answer describes all of the solutions to the desired equation.
Let $a+b=x^3$ and $c+d=y^3$ and $a+b+c+d=z^3$. Then $$x^3+y^3=z^3.$$ This is impossible for $z\neq 0$ by Fermat's Last Theorem (or special cases thereof which Fermat actually did prove). So, we need one of the summands to be zero.
Case 1: $a+b+c+d=0$
In this case, we get that both $a$ and $d$ and $a+d$ are perfect cubes since their negatives $b+c+d$ and $a+b+c$ and $b+c$. So, now we need that one of $a$ and $d$ and $a+d$ is zero.
Case 1a: $a=0$
Now, we get that $b$ is a perfect cube, since $a+b$ is. However, then one of $b$ and $c+d$ and $b+c+d$ must be zero. However, it must be that $b+c+d$ is zero, since $b=-c-d$ cannot be zero since $a$ is. At this point, we can solve the system $$a=0$$ $$b+c=\alpha^3$$ $$c+d=\beta^3$$ $$b+c+d=0$$ which has solutions for all $\alpha,\beta$. For instance $(0,-8,9,-1)$ is of this form.
Case 1b: $d=0$
This case, by symmetry, will yield the same result as the previous case, except with $(a,b,c,d)$ swapped to $(d,c,b,a)$.
Case 1c: $a+d=0$
Note that this implies $b+c=0$ as well, so $a=-d$ and $b=-c$ follow. Moreover, since $a+b+c=a$ is a cube, $a$ must be a non-zero cube. Then, we get the following family of solutions $$a=\alpha^3$$ $$a+b=\beta^3$$ $$c=-b$$ $$d=-a$$ which can be solved for all $\alpha,\beta$ input into the system. For instance, the tuple $(1,7,-7,-1)$ is of this form
Case 2: a+b=0
Since $a+b+c$ is a perfect cube, we get that $c$ is a perfect cube. It suffices to have $c$ and $b+c$ and $c+d$ and $b+c+d$ to all be cubes. In particular, if you have a solution to $$\alpha^3+\beta^3=\kappa^3+\gamma^3$$ then you can generate a solution as $$c=\alpha^3$$ $$b+c+d=\beta^3$$ $$b+c=\kappa^3$$ $$c+d=\gamma^3$$ Though more work is necessary to finish classifying this class of solutions.