Enumerating binary matrices by $X$-ray sequences Consider all $n\times n$ binary (entries are either $0$ or $1$) matrices, denoted $\mathcal{B}_n$.
Define the $X$-ray sequence of $A=(a_{ij})\in\mathcal{B}$ by $X(A)=x(1)x(2)\cdots x(2n-1)$ where
$x(k)=\sum_{i+j=k+1}a_{ij}$. Then, the number of distinct $X$-ray sequences can be easily seen to be $n!(n+1)!$.
Example. Let $A=\begin{pmatrix} 1&2&3\\3&4&5\\0&1&2\end{pmatrix}$. Then $X(A)=15762$.

QUESTION. If we specialize to the subfamily $\mathcal{F}_n\subset\mathcal{B}_n$ of such invertible (over the field $\mathbb{F}_2$) matrices, then is there a formula for the total number $u_n$ of distinct $X$-ray sequences? If this is asking too much, how about an asymptotic growth of such enumeration?

NOTE. The cardinality of $\mathcal{F}_n$ is $\prod_{j=0}^{n-1}(2^n-2^j)$.
UPDATE. I've now computed a few terms: $u_1=1, u_2=5, u_3=77, u_4=2150$.
 A: I asked in a comment the following: Is it true that if $A$ is a square
matrix over a field $K$, then there is a diagonal matrix $D$ for which
$A+D$ is nonsingular? Here is a proof.
Induction on $n$. Clear for $n=1$. Assume for $n$. Let $B=(b_{ij})$ be
an $(n+1)\times (n+1)$ matrix. Let $C$ be the submatrix indexed by
$1\leq i\leq n$, $1\leq j\leq n$. By the induction hypothesis, there
is a diagonal matrix $D=\mathrm{diag}(d_1,\dots,d_n)$ for which $C+D$
in nonsingular. Let $\alpha=\det(C+D)\neq 0$. Let
$D'=\mathrm{diag}(d_1,\dots,d_n,0)$. Expand $\det(B+D')$. The
coefficient of $b_{n+1,n+1}$ is $\alpha$. Since $\alpha\neq 0$, we can
add $c=0$ or $c=1$ to $b_{n+1,n+1}$ to get a nonzero
determinant. Hence $B+\mathrm{diag}(d_1,\dots,d_n,c)$ is
nonsingular. $\ \ \Box$
Going back to the original problem, we can specify arbitrarily the
entries off the main antidiagonal and then choose an antidiagonal
making the determinant nonzero (by row and column permutations, it
makes no difference here that we are looking at antidiagonals rather
than diagonals),  giving at least $n!^2$ X-ray sequences.
