Enumeration of $0-1$ matrices with determinant $1$ Has the number $f(n)$ of $n \times n$, $0{-}1$ matrices whose determinant is $+1$
been enumerated?
E.g., for $n{=}2$, there are $f(2)=3$ such matrices:
$$
\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 1 \\
\end{array}
\right)
\;,\;
\left(
\begin{array}{cc}
 1 & 0 \\
 1 & 1 \\
\end{array}
\right)
\;,\;
\left(
\begin{array}{cc}
 1 & 1 \\
 0 & 1 \\
\end{array}
\right)
\;.
$$
For $n{=}3$, I count $f(3)=84$ such matrices,
from
$$
\left(
\begin{array}{ccc}
 0 & 0 & 1 \\
 1 & 0 & 0 \\
 0 & 1 & 0 \\
\end{array}
\right)
\;,
$$
to
$$
\left(
\begin{array}{ccc}
 1 & 1 & 1 \\
 1 & 1 & 0 \\
 0 & 1 & 1 \\
\end{array}
\right)
\;.
$$
These matrices are a subset of $SL(n,\mathbb{R})$.

Update. Oh, I see $f(n)$ is OEIS A086264:
$$
1, 3, 84, 10020, 4851360, 9240051240.
$$
No substantive information is provided in OEIS besides those six
computed values.

Addendum.
Unrevealing, but just as a curiosity, here is an overlay of the $84$ 
equal-volume parallelepipeds
that result by applying the $n{=}3$ matrices to the $3 \times 2 \times 1$ box with
lowerleft corner at the origin:
 

 A: Will Orrick might have a good guess for this one.  As far as I know the answer has only been determined for n up to 8.  The number of matrices with odd determinant is known: it is $$\prod_{i=0}^{n-1}(2^n - 2^i)$$, which is about $0.3 * 2^{n^2}$.  Noam Elkies has the best guess, but since the number of matrices achieving larger determinants drops off rapidly, I would guess more like $2^{n^2 -cn}$ for a small positive value of $c$.
From the arxiv paper of Zivkovic in a comment above, one has for matrices with absolute determinant value 1 : n=6, 18480102480; n=7,135491563468800; n=8, 3766962568171582080 .
It is foolish to conjecture a value for $c$ based on this small amount of data, so I will
only guess that $c \in [1/2 , 1]$. 
One can use certain methods to guess at a better lower bound.  I like the adjunction method of adding a row and column to a matrix of ADV 1.  This gives a lower bound for $a_{n+1}$ of roughly $n^2*2^n*a_n$, with $a_n$ being the number of matrices of ADV=1 and order $n$.  
Gerhard "It's Nice To Be Referenced" Paseman, 2015.03.23
