There is a lot to unpack here, so let me try to go line by line:
We know that the volume of higher dimensional sphere is inversely proportional to the number of dimensions.
This isn't quite right. First off, I think that you mean the unit ball (not the unit sphere). The sphere is the set
$$ S^n(0,r) := \{ x \in \mathbb{R}^n : \|x\| = r \}, $$
whereas the ball is the set
$$ B^n(0,r) := \{ x \in \mathbb{R}^n : \|x\| < r \}. $$
Essentially, the ball is solid, while the sphere encloses the solid ball. Typically, when we use the phrase "volume", we are talking about balls, not spheres. I am going to assume that by "sphere," you actually mean the ball.
Note, also, that the volume of the unit ball $B^n(0,1)$ tends to zero as $n$ goes to infinity, but it is not quite correct to say that the volume is inversely proportional to the dimension. This implies that there is some constant $k$ such that
$$ \operatorname{vol}(B^n(0,1)) = \frac{k}{n}. $$
No such constant exists. It is actually worse than that.
The volume of the ball $B^n(0,r)$ is given by
$$ \operatorname{vol}(B^n(0,r)) = \operatorname{vol}(B^n(0,1)) r^n = \frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2} + 1)}r^n, $$
which (clearly?) tends to zero with $n$ if $r \le 1$. It is less clear (perhaps) that this tends to zero if $r > 1$, but $\Gamma(\frac{n}{2}+1)$ is roughly $\lceil\frac{n}{2}\rceil!$ for large $n$, which grows faster than the numerator. In any event, I think that we might want to restrict our attention to the unit ball.
Hence, as the we increase the number of dimension keeping the radius fixed, the volume of the sphere tends to Zero.
This is almost right. If we rewrite this sentence to read "As we increase the dimension, the volume of the unit ball tends to zero," it will be correct.
On the other hand, if we increase the number of dimension in case of Cubes or polyhedra (to be specific), the volume increases. Hence, the volume of an infinite dimensional hypercube is infinite.
The volume of a cube, which I take to be the set
$$ C^n(0,r) = [-r,r]^n, $$
is given by
$$ \operatorname{vol}(C^n(0,r)) = (2r)^n. $$
This tends to infinity with $n$ if and only if $r > \frac{1}{2}$, and tends to zero if and only if $r < \frac{1}{2}$. This is basically the same behaviour as the ball, with one little caveat:
$$ \lim_{n\to\infty} \operatorname{vol}(C^n(0,1/2)) = 1 \ne 0 = \lim_{n\to\infty}\operatorname{vol}(B^n(0,1)). $$
Hence we are, perhaps, most interested in the unit cube, as well.
Now, in the context of topology, a sphere is homeomorphic to a cube or parallelopiped,
This is definitely true in finite dimensional spaces. It is not obvious that it is true in infinite dimensional spaces. Basically, there are many norms that we can define on $\mathbb{R}^n$, two of which are coming into play here. The first, called the Euclidean norm, is given by
$$ \|(x_1,\dotsc,x_n)\|_2 := \sqrt{x_1^2 + \dotsb + x_n^2}. $$
The second, called the $\ell^\infty$ or maximum norm is given by
$$ \|(x_1,\dotsc,x_n)\|_\infty := \max\{|x_1|,\dotsb,|x_n|\}. $$
Both of these norms induce topologies on $\mathbb{R}^n$, and it turns out that if $n$ is finite then these topologies are equivalent (they have the same open sets). The open balls for the Euclidean norm are the usual open balls, and the open balls for the max-norm are the open cubes, which (after some abstract nonsense) implies that cubes and balls are homeomorphic in finite dimensional spaces.
However, the Euclidean norm and the max-norm do not induce the same topology on $\mathbb{R}^\infty$. Much more delicate arguments need to be made, and it is not at all clear to me that a cube and ball in $\mathbb{R}^\infty$ are homeomorphic (or even what the right topology is).
in the case of infinite dimensions... where the volume of the sphere is infinite and the volume of the hypercube is infinite...
I think that I have already addressed this point.
Does that imply that an object of volume of the magnitude Zero is homeomorphic (that it can be continuously transformed) to something which has the volume of the magnitude Infinity?
Basically, I think that you are asking if an object with infinite volume can be homeomorphic to an object with finite volume, or even zero volume. This is not a surprising result. For example, the unit disk in $\mathbb{R}^2$, i.e. the set $B^2(0,1)$, has "volume" (area, measure, whatever you want to call it) $\pi$. However, the unit disk is homeomorphic to all of $\mathbb{R}^2$, which has infinite volume (area, measure, whatevs).
The issue here is that topologies are very weak structures. The tell us how points relate to each other. Topologies give a very weak notion of "nearness", allow us to define limits, give us a way of talking about connectedness (several ways, in fact), and other such structures. However, there is no intrinsic way to measure volume, and there is no reason to believe that volume should be a topological invariant (i.e. two homeomorphic objects should have the same volume).
If you want to talk about volumes, you need more structure than just a topology. Typically, we talk about volumes in a part of mathematics called measure theory. In this theory, the measure of a set can potentially communicate with the topology (this is, essentially, the idea of a Borel measure on $\mathbb{R}^n$), but the measure gives us more information than the topology alone. Measure spaces have "stronger" structures than topological spaces.
