Define a ground set $X$ of size $2^{n-1}-1$.  Now choose $2^{n-1}-1$ subsets of $X$, each of size at least 2, such that the sum of their sizes is $(n-2)2^{n-1}+2$ (so the average size is slightly more than $n-2$), and such that no two sets overlap in more than one element.  

Now take the poset consisting of the empty set, the singletons, the chosen subsets, and the full set $X$, ordered by inclusion.  Because no two sets overlap in more than one element, the join of any two singletons in well-defined, and it follows immediately that the join is always well-defined.  Since the poset is also bounded, it is a lattice.  By construction, it has $2^n$ elements and $n2^{n-1}$ edges.  

We now go back and check for what $n$ it will be possible to choose a set system such as we have described. This isn't hard.  For odd $n$, we can consider our ground set to be non-zero points in the plane $\mathbb F^2_{2^{(n-1)/2}}$.  There are more than $2^{n-1}$ affine lines in this plane, no two intersect in more than one point, and each contains at least $2^{(n-1)/2}-1$  points of our ground set.   (If the sum of the sizes of our sets is too big, we can throw away elements from the sets.  The intersection property will obviously be preserved.)  Since $2^{(n-1)/2}-1$ is larger than $n-1$ if $n$ is at least 7, we are certainly done for all odd $n$ at least 7.  A one-off construction for $n=5$ seems easy enough.  Even $n$ should also be straightforward.  I don't know whether or not my construction works for $n=4$, but it should have no trouble starting at $n=6$.