If there are plane filling curves, are there volume-filling planes? Not a mathematician, so don't know if this question even makes sense.
But if you can tessellate a plane (which is 2d) with a line (1d), can you extend the concept to higher dimensions?
How would that look like?
 A: Usually a space-filling curve is understood to be a continuous surjection $u : I \to I \times I$ (where $I = [0,1]$). If you want, you can use $u$ to construct a continuous surjection $v : \mathbb{R} \to  \mathbb{R}^2$.
An immediate consequence is that we can find continuous surjections  $u^m_n : I^m \to I^n$ whenever $m \le n$ (similarly $v^m_n : \mathbb{R}^m \to  \mathbb{R}^n$).
To see this, we start with $m = 1$ and define $u^1_n$ inductively by $u^1_1 = id$ and
$$u^1_{n+1} : I \stackrel{u^1_n}{\rightarrow} I^n = I^{n-1} \times I \stackrel{id \times u}{\rightarrow} I^{n-1} \times I^2 = I^{n+1} .$$
If $p_m : I^m \to I$ denotes projection to the first coordinate, we set
$$u^m_n = u^1_n \circ p_m .$$
In your question you use the wording "tessalation". Usually this is a tiling of a plane using one or more geometric shapes ("tiles") such that no point is contained in the interior of more than one tile. You can generalize this concept to higher dimensions, but you must be aware that space filling curves do not really have a relation to tessalations in that sense.
