It would seem that ListVectorPlot only works for vector arrays of at least 3 by 3. For smaller arrays I get the message VisualizationCoreListVectorPlot::vfldata

With[{n = 2}, 
 ListVectorPlot[Table[{1, 1}, {n}, {n}], VectorPoints -> All]]

I wonder why the message does not appear with your input

ListVectorPlot[{{{1, 2}, {3, 4}}, {{5, 6}, {7, 8}}}]

If you want as many vectors as data, just use the option VectorPoints->All:

With[{n = 3}, 
 ListVectorPlot[Table[{1, 1}, {n}, {n}], VectorPoints -> All]]
With[{n = 3}, ListVectorPlot[RandomReal[{0, 1}, {n, n, 2}]]]

As Jim Baldwin points out, the first data set has insufficient rank to form a basis for interpolation. In version 10 this case did generate an error message, which no longer appears in 11. The reasonable solution would be to use the option VectorPoints->All, which tells ListVectorPlot to just plot the vectors as given -- no need for interpolation.

Unfortunately, even with that option, ListVectorPlot and ListVectorPlot3D still require a full rank set of vectors suitable for interpolation. I say unfortunately, because every time I have had occasion to use one of these functions I have wanted it to plot the vectors given -- I have never had a use for the interpolation, which I could construct myself if needed.

I reported this about 18 months ago, with the suggestion that VectorPoints->All eliminate the requirement for a full rank data set, but no joy so far.