Let's define $k$-blocking set in affine space $AG(n,q)$ a set that meets every coset (translate of subspace) of dimension $k$. 


I have seen a lot work related to minimal $(n-1)$-blockings set.


[Covering finite fields with cosets of subspaces](http://www.sciencedirect.com/science/article/pii/0097316577900012). 


[The Blocking Number of an Affine Space](http://ac.els-cdn.com/0097316578900134/1-s2.0-0097316578900134-main.pdf?_tid=a4307ca4-80ca-11e5-b438-00000aab0f6c&acdnat=1446404533_eaabb01608cd39e780dbbd0e451260cb) 


In these articles it is proved that minimal $(n-1)$-blocking has $n(q-1)+1$ points. 


I can't find any result about minimal $(n-2)$-blocking sets (even for $AG(n,2)$). I have managed to prove following bounds about $(n-2)$-blocking set in $AG(n,2)$. It has at least $2n-1$ and no more than $3n^{\log\_{2} 3}+1$ points.


Here is a link to my paper.
<http://ysu.am/files/8.On%20The%20Minimal%20Coset%20Covering%20of%20Solutions%20of%20a%20Boolean%20Equation.pdf>


I am very interested in the solution of the problem. Does anyone have any information about it?


I have also asked following questions related to problem:
<https://math.stackexchange.com/questions/869308/blocking-set-for-cosets-of-codimension-2>
<https://math.stackexchange.com/questions/863592/find-minimal-set-of-cosets-c-so-that-each-2-vectors-in-a-n-are-in-one-cos>



$(n-2)$-blocking sets in $AG(n,2)$
**Not much is known for the general case.** 


Let $m(k, n, q)$ denote the minimum size of an $k$-blocking set in $AG(n, q)$. Trivially we have $m(0, n, q) = q^n$ and $m(n, n, q) = 1$. By Jamison/Brouwer-Schrijver we get $m(n-1, n, q) = 1 + n(q-1)$ as you have mentioned. To at least give bounds on other values we can prove the following inequality $$qm(k, n-1, q) \leq m(k, n, q) \leq (q-1)m(k, n-1, q) + m(k-1, n - 1, q).$$


The lower bound follows from taking $q$ parallel hyperplanes that cover the space and then observing that the intersection of the blocking set with each of these hyperplanes is $k$-blocking set in $AG(n-1, q)$. For the upper bound, again take $q$ parallel hyperplanes, $H\_1, \dots, H\_q$. For $1 \leq i \leq q - 1$, let $B\_i$ be a $k$-blocking set in $H\_i$ and let $B\_q$ be a $(k-1)$-blocking set in $H\_q$. Then one can argue that $B = B\_1 \cup \dots \cup B\_q$ is a $k$-blocking set in $AG(n, q)$ as follows. If $S$ is a $k$-flat in $AG(n,q)$ which is contained in any $H\_i$ for $i < q$, then it is blocked by $B\_i$. If it is contained in $H\_q$, then it is blocked by $B\_q$ since $B\_q$ blocks all $(k-1)$-flats, it must block all $k$-flats as well. If $S$ is not contained in any of the $H\_i$'s then its intersection with $H\_q$ is a $(k-1)$-flat, which is again blocked by $B\_q$. 


For your particular case of $(n-2)$-blocking sets, this tells us that $$q(1 + (n-1)(q-1) \leq m(n-2, n, q) \leq (q-1)^2\binom{n}{2} + (q-1)n + 1.$$


So, for a constant $q$ there is a linear lower bound and a quadratic upper bound. I don't know if there has been any improvement ``in general'' besides the one that you mention for $q = 2$. (Do your methods extends to $q > 2$? If yes, then do they improve these bounds?) I have asked a couple of colleagues who work in finite geometry and they don't know either. 


For some particular values, note that in $AG(3,3)$ the complement of such a blocking set is a cap (see <https://www.mathi.uni-heidelberg.de/~yves/Papers/CapSurvey.pdf>). And we know that the maximum size of a cap in $AG(3,3)$ is $9$. Thus, $m(1, 3, 3) = 18$, which is one less than the upper bound we get above. I guess some other specific values can also be computed. Similarly, the problem of finding smallest 1-blocking sets in $AG(n,3)$ is equivalent to finding the largest caps in $AG(n,3)$, on which there is significant progress in general. See [New bounds on cap sets](http://www.ams.org/journals/jams/2012-25-02/S0894-0347-2011-00725-X/). 


Note that the problem is completely solved for projective spaces. We know that a $k$-blocking set in $PG(n,q)$ has size at least $\big[ {n-k + 1 \atop 1} \big]\_q = 1 + q + \dots + q^{n-k}$ with equality if and only if the blocking set is an $(n-k)$-dimensional subspace. But if you want to talk about ``minimal'' blocking sets instead of the smallest ones, then the problem again becomes much harder. 


For a recent survey on projective blocking sets see Chapter 3 in [current research topics in galois geometry](https://www.novapublishers.com/catalog/product_info.php?products_id=21439). There is also a small section on affine blocking sets where they mention the same recursive bound. They follow a different terminology though: what you have called an $(n-k)$-blocking set is called a $k$-blocking set. 


**Edit** When $n = 3$ there are some nice improvements to the bounds. Peter Sziklai, and later Simeon Ball proved general results on $1$-blocking sets in affine spaces, which imply that the size of such a set in $AG(3,q)$ is at least $2q^2 - 1$. See [Blocking sets in three dimensional finite affine spaces](https://mathoverflow.net/questions/229417/blocking-sets-in-three-dimensional-finite-affine-spaces) for the details. Also, the lower bound can be improved in general to $m(n-2, n, q) \geq (n-1)(q^2 - 1) + 1$. See Section 3 in <http://www-ma4.upc.es/~simeon/polynomialmethod.pdf> 


