the answer is $2n+1$. first of all it is clear that at least this many are required because no $1\times 2n+1\times 2n+1$ can be completely covered without the small cubes. this is because the second type of bar only intersects such a plane in even number of cubes. now for the example we can inductively fill the $2n-1\times 2n-1\times 2n-1$ cube in the middle using $2n-1$ small cubes. for the "perimeter" we can put a $2k\times 2k \times 1$ bar on each side such that only two opposing corners are left empty (think of a rose petal), we fill the two remaining spaces with small cubes. We have used $2+2n-1=2n+1$ small cubes thus finishing the induction.