What fraction of n x n invertible integer matrices contain at least one unit? The question is simple:

What fraction of matrices in $G_n = \text{GL}_n(\mathbb{Z})$ have at least one unit entry (i.e., either $\lbrace\pm 1 \rbrace$)?

I'm not sure what the correct measure  on $G_n$ would be, so here is a suggestion: for each natural number $m \geq 1$, define $G_n(m)$ to be the set of precisely those matrices in $G_n$ whose entries are bounded in absolute value by $m$ and let $H_n(m)$ be the subset of matrices which have at least one unit entry. These are finite and non-empty sets, so in particular for each $m$ the following ratio is defined: $$r_n(m) = \frac{|H_n(m)|}{|G_n(m)|}$$
Now, one can take limits (or lim-sups?) as $m \to \infty$. Again, this is only a suggested measure and it should not constrain potential answers: all reasonable measures are welcome.
Motivation
I write software that pre-processes large (filtered) cell complexes via discrete Morse theory to produce smaller cell complexes with identical homology groups. Without getting into gory details, the basic idea is to greedily exploit unit incidence among cell-pairs in order to clear out the corresponding row and column from the matrix representation of a boundary operator via obvious row and column operations: once these have been cleared, these paired cells can be removed from the complex altogether.
Recently, I was handed a collection of triangulated homology $4$-spheres with tons of torsion in the fundamental groups. On these complexes, the naive greedy collapsing schemes do not produce a perfect reduced complex (i.e., with one zero-dimensional cell and four dimensional cell). In fact, the boundary matrices of the reduced complexes often contain no units at all, and this is precisely when no more collapses are possible. I would like a quantification of how often should one expect an invertible integer matrix to have exploitable units? in order to judge the performance of discrete Morse theoretic reductions on these spheres.
 A: I imagine the answer is that the limiting probability equals 0.
Even the asymptotics for $|G_n(M)|$ are nontrivial. See example 1.6 in Duke/Rudnick/Sarnak. I would suggest looking at one of Shparlinski's papers and the references cited therein; maybe what you need has already been done.
My hope would be that you could count the number of matrices in $|G_n(M)|$ with a prescribed fixed first row, say. If so, you could solve this problem just by considering the 1s in the first row.
A: Using arguments like those indicated by smoked-salmon-sandwiches, it will follow that the "probability" $|H_n(m)|/|G_n(m)|$ tends to zero exponentially fast with the height $m$. This should be a special case of the general results of Gorodnik and Nevo (see their book in Annals of Math. Studies). 
Another way of measuring things with the same conclusion is to take a generating set $S$ of $\mathrm{GL}_n(\mathbf{Z})$ and consider a random walk using $S$ (and the inverses of elements of $S$).  After $k$ steps, it will transpire that the probability of being in the subset $H_n$ will be $\leq c_1c_2^{-k}$ for some $c_1>0$, $c_2>1$.  Here the argument is a bit more transparent maybe: basically, the equidistribution modulo a prime follows from elementary Markov chain methods, and the uniformity over primes that implies exponential decay (by choosing a suitable prime) is a direct consequence of Kazhdan's Property (T) (for $n\geq 3$) or Property ($\tau$) for $n=2$.
Arguments like this are known as "escape from subvariaties" in the recent literature concerning sieve in discrete groups (though the focus there is on more complicated counting problems.)  There is some discussion in the paper "Affine linear sieve" of Bourgain, Gamburd and Sarnak and in recent surveys of this topic which can be found by googling around.
A: Really a couple of comments on @smokedsalmonsandwich's answer:
for the result $\mod p$ the sledgehammer way of dealing with this is the Lang-Weil bound: restricting some specific entry to 1 defines a proper sub variety of algebraic group $SL(n)$ and so for large $p$ the number of restricted matrices is like $c/p$ times the number of unrestricted matrices. You are in the union of $n^2$ such subvarieties, so you get something very similar to @smokedsalmonsandwich's bound.
Secondly, if the matrices have different restrictions on coefficients, there is relevant work of Ahmadi-Shparlinsky: Distribution of matrices with restricted entries over finite fields
O Ahmadi, IE Shparlinski - Indagationes Mathematicae, 2007
Thirdly, one can get sharp error bounds on the asymptotics restricted to congruence subsets using Nevo/Sarnak and Gorodnik/Nevo.
Fourthly, using all of the above together with sieve machinery (see, e.g., Emmanuel Kowalski's book) should give decent bounds on what the odds are of finding a matrix with a $\pm1$ entry.
A: Let $G_p$ denote the subgroup $\mathrm{GL}_n(\mathbf{F}_p)$ consisting
of matrices with determinant $\pm 1$. Then
$G_p$ is exactly the image of $\mathrm{GL}_n(\mathbf{Z})$ under reduction mod $p$.
Any natural method of counting matrices of "height at most $T$" should have the following property: if one restricts to matrices satisfying some congruence condition
corresponding to some subset $S_p \subset G_p$, then the asymptotics should be modified by the factor $|S_p|/|G_p|$.
On the other hand, as $p \rightarrow \infty$, the number of elements in $G_p$ with
an entry in $\pm 1$ goes to zero. Here is an easy proof, which shows
that the probability is at most $2 n^2/(p-1)$, if $n \ge 2$. 
Permuting the rows and columns preserves $G_p$. Hence the probability that any particular fixed entry is $\pm 1$ is equal to the probability that the first entry is $\pm 1$. 
Hence the probability that any entry is $\pm 1$ is at most $n^2$ times the probability
that any fixed entry is $\pm 1$. Since $n \ge 2$, $G_p$ contains the diagonal matrix
with terms $\{\epsilon, \epsilon^{-1}, 1, 1, \ldots, 1\}$, where $\epsilon$ is a primitive root. Multiplication by the $k$th power of element gives a bijection between terms whose first entry is one with terms whose first entry is $\epsilon^k$. Hence the probability that the first term is $\pm 1$ is $2/(p-1)$ times the probability it is non-zero (which is obviously at most $1$).
Hence the "probability"
that any term is $\pm 1$ is asymptotically at most $2n^2/(p-1)$ for any $p$, and hence $0$.
It remains to show that the "natural" forms of counting do satisfy this hypothesis.
If one counts columns by their Euclidean norm, then, in this case, the result follows
from work of
Borovoi and Rudnick:
http://www.math.tau.ac.il/~borovoi/papers/hardy.pdf
