I believe it is possible to use some recent closely related work of Kenneth Maples to get a much better (but probably still not quite tight) bound.  Let $C>0$ be a constant to be chosen later.  Call an $n \times n$ matrix $A$ *good* if it satisfies each of the following properties

P1. $A$ is non-singular over $\mathbb{R}$.

P2. $|det(A)|$ has at most $C \log n$ prime factors.

P3. $A$ has rank at least $n-C \log n$ over ${\mathbb F}_p$ for every prime $p$.

Here are two claims which together would imply $n+O(\log n)$ vectors are enough.

Claim 1: A random $n \times n$ $(0,1)$ matrix is good with probability $O\left(C^{-1}\right)$.

Claim 2: If $A$ is any fixed good matrix, augmenting $A$ by $5C \log n$ random rows with high probability leads to a matrix whose rows span ${\mathbb Z}^n$.

We first look at claim 1.  The probability P1 fails is exponentially small in $n$ (as originally shown by Kahn, Komlos, and Szemeredi).

For P2 and P3, we use the following result of Maples (Corollary 1.3 [here][1]): For any prime $p$, the probability that a random $n \times n$ matrix has rank $n-k$ over ${\mathbb F}_p$ is

$$p^{-k^2} \frac{\prod_{\ell=k+1}^{\infty} \left(1-p^{-\ell}\right)}{\prod_{\ell=1}^k \left(1-p^{-\ell}\right)}+O\left(e^{-cn/2}\right),$$
where both $c$ and the constant implicit in $O()$ are independent of $p$.
We can actually bound the probability above by $O\left(p^{-k^2} +e^{-cn/2}\right)$, since the ratio of products is at most $\prod_{\ell=1}^{\infty} (1-2^{-\ell})^{-1}$.  Summing over all $k$, the probability $A$ is singular over ${\mathbb F}_p$ is $O\left(\frac{1}{p} + e^{-\frac{cn}{2}}\right)$.  Summing over all $p$, the expected number of primes less than $e^{cn/4}$ dividing $|det(A)|$ is at most $\log n +O(1)$.

There can be at most $2 \log n/c$ prime factors of $|det(A)|$ larger than $e^{cn/4}$, since otherwise $|det(A)|$ would be larger than $n^{n/2}$ and violate Hadamard's bound.  So the total expected number of factors is $O(\log n)$, and the probability P2 fails is $O(1/C)$ by Markov's inequality.

For P3, we again split into small and large primes.  Applying Maples' theorem again, the probability P3 fails for a given prime less than $e^{cn/4}$ is at most $O\left(p^{-C^2 \log^2 n}+ e^{-cn/2}\right)$, and by the union bound the probability P3 fails for some small prime is small.

For large primes, we use the observation that $A$ can only have rank less than $n-k$ over ${\mathbb F}_p$ if $p^k$ divides the determinant of $A$ (e.g. because in this case we can row reduce over the integers so $k$ rows have all entries divisible by $p$, at which point we can pull a factor of $p$ out for each row).  In particular, if $C$ is sufficiently large we know from Hadamard's bound it is impossible for P1 to succeed and P3 to fail for some prime larger than $e^{cn/4}$.   This finishes Claim 1.

We now turn to Claim 2.  We first note that the for $m \geq n$ the vectors $v_1, \dots, v_m$ span ${\mathbb Z}^n$ if and only if the matrix with the $v_i$ as rows has full rank over ${\mathbb F}_p$ for every prime $p$ (if the volume of a cell is $V$, then $V$ divides the determinant of every $n \times n$ submatrix).  Since $A$ is good, we know that we already are full rank for all but at most $C \log n$ primes.  So it is enough to show the augmentation with high probability fixes each of those primes.  Fix any one such prime $p$.

We use the following observation (originally due to Odlyzko): Any proper subspace of ${\mathbb F}_p^n$ contains at most half of the $(0,1)$ vectors (e.g. because if you fix a column basis, whichever column is not in the basis is determined by the remaining $n-1$ columns).  It follows that so long as $v_1, \dots ,v_j$ do not already span the space,
$$P\left(v_{j+1} \notin Span(v_1, \dots v_j)  \right) \geq \frac{1}{2}.$$
By assumption P3, $A$ already had rank at least $n-C \log n$ before we added the rows.  The only way $A$ can fail to be full rank after the augmentation is if the above event occurred at least $4 C \log n$ times, an event which occurs with probability at most
$$\binom{5C \log n}{4 C \log n} 2^{-4C \log n} = 2^{(-0.39+o(1)) C \log n}.$$
Taking the union bound over all $p$ which divide $|det(A)|$, the probability we fail to be of full rank modulo some prime is at most $C \log n 2^{-(0.39+o(1)) C \log n} = Cn^{-0.39C+o(1)}$, proving Claim 2.

This bound is probably still not quite tight, especially in the handling of P3 for large $p$. One annoyance in trying to drop below $\log n$ is that if the last row of $A$ and all the rows added in the augmentation are zero (an event occurring with probability roughly $2^{-n(m-n)}$), the matrix fails to be of full rank modulo every prime.  This means just taking the union bound over all the roughly $2^{c n \log n}$ primes less than $n^{n/2}$ won't be enough if $m-n$ is much smaller than $\log n$, unless we could possibly get some handle on the event "$A$ is of full rank over  $\mathbb{R}$ but not over ${\mathbb F}_p$" for large $p$. 

  [1]: http://user.math.uzh.ch/maples/maples-cokernel.pdf