Here are some answers, but let me introduce some notation: What you are calling the *norm* of $x\in\mathbb{C}{\otimes}\mathbb{O}$, I will denote by $N(x) = x\bar x = \bar x x\in \mathbb{C}$. The complexified octonions satisfy $N(xy) = N(x)N(y)$, and the norm is a nondegenerate quadratic form on $\mathbb{C}{\otimes}\mathbb{O}$, considered as a complex vector space.

1) No. Given a zero divisor $u$, consider the quadratic form $Q_u(x) = N(xu)$ on $\mathbb{C}\otimes\mathbb{O}$. This quadratic form is identically zero, so this implies that the subspace $(\mathbb{C}{\otimes}\mathbb{O})u$ is a totally $N$-null linear subspace of $\mathbb{C}{\otimes}\mathbb{O}$, an $8$-dimensional vector space on which $N$ is nondegenerate. Thus, this subspace has dimension at most $4$ and it contains all of the elements $xu$ where $x$ is invertible. Thus, these latter elements can't span $\mathbb{C}{\otimes}\mathbb{O}$ all zero divisors because the cone $N(u) = 0$ of zero divisors does not lie in any proper linear subspace of $\mathbb{C}{\otimes}\mathbb{O}$. A similar argument applies to the elements $ux$ where $x$ is invertible.

2) Yes. To understand this, you need to realize that there are two kinds of zero divisors. The first kind (generic) are the zero divisors $u$ such that $u+\bar u = \lambda\not=0$, i.e., $u$ is not 'purely imaginary'. For such a $u$, we have $$ 0 = u\bar u = u(\lambda-u) = \lambda u - u^2,
$$
so setting $a = u$ and $b = \bar u/\lambda$, we have $u = a\bar b$. The second kind has $\bar u = - u\not=0$, so we can write $u = p + iq$ where $p$ and $q$ are purely imaginary elements of $\mathbb{O}$ that have the same norm, say, $p\bar p = q\bar q = r>0$, and are orthogonal. Then $1,p,q,pq$ span an associative quaternion subalgebra of $\mathbb{O}$. We then compute, using that $p^2 = q^2 = -r$ and $pq=-qp$ that
$$
(1+ipq)(p+iq) = (1+r)(p+iq).
$$
Dividing by $1+r>0$, we have $p+iq$ written as a product of zero divisors.

3) For any (nonzero) zero-divisor $a$, the subspace consisting of those $b$ that satisfy $ab = 0$ is a (complex) vector space of dimension $4$. Hence the subspace consisting of (left) multiples of $a$ has dimension $8-4 = 4$ as well. The same dimension counts hold true for right multiplications. In particular, by dimension count, the subspace consisting of elements of the form $\bar a x$ is equal to the subspace consisting of elements $b$ that satisfy $ab=0$.

4) As a complex analytic variety, it has dimension 11. The reason is that, in the pair $(a,b)$, as a fixed $a\not=0$ ranges over the $N$-null cone (which has complex dimension $7$), $b$ is only required to lie in the $4$-dimensional subspace that is the kernel of left multiplication by $a$. When $a=0$, you get an $8$-plane, since there is then no restriction on $b$, but this doesn't affect the count.

5) (*Corrected after the OP's comment.*) The complex dimension of this singular variety is $14$. The reason is that the set of pairs $(a,c)$ that satisfy $ca=0$ has complex dimension $11$, and, for each such pair with neither $a$ nor $c$ equal to zero, the equations $ab=bc=0$ are linear equations for $b$ that, it turns out, have a (complex) $3$-dimensional space of solutions. Thus, the variety has dimension $11+3 = 14$. (The special cases when one of $a$, $b$, or $c$ vanishes only have dimension $11$, so they can't increase the count.)