Compute the 4-dimensional volume of the set $\ E\ \subset \mathbb{R}^{4}\ $ consisting of all$\ (x,y,z,t) \in \mathbb{R}^{4}\ $ such that
$$\left|x\right|+\left|x+2y\right|+\left|x+2y+3z\right|+\left|x+2y+3z+4t\right|\ \leq a\ $$ where $\ a > 0\ $ is a given number.
My question is: can i consider the relation
$\left|x\right|+\left|x+2y\right|+\left|x+2y+3z\right|+\left|x+2y+3z+4t\right|\ \leq a\ $
to be the same as:
$\ {x}^{2}+{(x+2y)}^{2}+{(x+2y+3z)}^{2}+{(x+2y+3z+4t)}^{2} \leq a?\ $
or could it also be equivalent to:
$\ {x}+{(x+2y)}+{(x+2y+3z)}+{(x+2y+3z+4t)} \leq a?\ $
Otherwise, I am not sure how to deal with the case of absolute value or presence of norm when computing volume.
No, it is not true. The sum of absolute values gives you a polyhedron while the sum of squares will give you a round object. If you work in $2D$, you find that $|x|+|y|= a$ is a square oriented on a diagonal, while $x^2+y^2= a^2$ is a circle of radius $a$.
You are expected to set up a four dimensional integral. As all the terms are positive, the first tells you that the $x$ integral runs from $-a$ to $+a$. Now find the range of $y$ that corresponds to a given $x$. That will give you the range of the $y$ interval. Keep going.