How many unit simplices are needed to cover a unit $n$-cube? The volume of an $n$-dimensional simplex of unit edge length is
$$V(n) = \frac{\sqrt{n+1}}{n! 2^{n/2}} \;,$$
so at least $\lceil 1/V(n) \rceil$ such simplices are needed to cover the unit $n$-cube.

Q1. What is an upperbound on the number of unit simplices needed to cover a unit $n$-cube?

I suspect that the $1/V(n)$ lowerbound is weak; 
perhaps there is no $c>0$ such that $c/V(n)$ is an upperbound?
For $n{=}2$, $1/V(2)=4/\sqrt{3} \approx 2.3$, but $4$ equilateral triangles are needed.
For $n{=}3$, $1/V(3)=6\sqrt{2} \approx 8.5$; 
I don't know how many tetrahedra are needed to cover the cube.

Q2. How many unit tetrahedra are needed to cover a unit cube?

 A: A cube of side $1/\sqrt{2}$ can be covered by $5$ unit tetrahedra (one using four of the cube's vertices, and the reflections of that one across each of its faces).
$8$ of those cubes can cover the unit cube, so we need at most $40$ tetrahedra that way.
A: I'm now thinking the answer is 22, or very close to it.  Put tetrahedral caps on each corner of the
unit cube,
and reflect about the base to place a corresponding interior tetrahedron, accounting for
16.  After observing how an interior tetrahedron maximally intersects a corner cube of
length 1/2, we see remaining 
6 square pyramids (one on each face of the unit cube) of height less than 1/2 and length of base
sqrt(2) - 1.  I'm thinking each will fit inside a unit tetrahedron.  Even if I am wrong, such
a pyramid should fit into a union of two tetrahedra, giving an upper bound of 28.
A: Following the suggestion of TMA just for Q2,
I found a calculation and beautiful graphics by
Anders Kaseorg
at this 
website

     


showing that a cube of sidelength $x \approx 0.2959$ fits inside a unit tetrahedron
in $\mathbb{R}^3$.
Because $1/x \approx 3.4$, a $4 \times 4 \times 4 = 64$ arrangement of cubes
cover a unit cube (with much to spare).
So a unit cube can be covered by $64$ unit tetrahedra.
Although there is considerable over-coverage (see below), 
a straightforward $3 \times 3 \times 3$ stacking does not suffice.

     


$64$ is surely much larger than the optimal.
A: From the upper bound $\sqrt{2}n$ on the edge length of the smallest regular simplex containing the unit cube in $\mathbb{R}^n$, shortly outlined in the question Smallest regular simplex containing the unit cube in $R^n$, it follows that roughly $n^n2^{n/2}$ unit simplices are sufficient to cover the unit cube by translation.
This might be far from optimal, since the translation covering density of any convex body in $\mathbb{R}^n$ is only $n\ln{n} + n\ln\ln{n} + 5n$, due to Rogers. Moreover, it is known that the most economical covering cannot be lattice like. (For the details and references, please let me refer to Section 1.3 in P. Brass, W. Moser, J. Pach, Research problems in discrete geometry, 2005.)
