It is not true in general. Let $n=3$, and take a piece of a brick wall (thickness one brick) that is homeomorphic to a 3-disc. For example, just take a largish rectangular wall. There are bricks of four colours, in multiple rows. In each even row the bricks are alternately colours $a,b,a,b,\ldots$ and in each odd row the bricks are alternately colours $c,d,c,d,\ldots$. The four open sets are the bricks of a given colour together with some surrounding mortar. Because of the way that brick walls are built, each 3-fold intersection is non-empty but the 4-fold intersection is empty.
Note that each of the open sets has lots of components, as do all of the 2- and 3-fold intersections. For counterexamples you will need the open sets and their intersections to be far from being contractible, so that the disc that is their union is far from being the nerve of the covering.
This answer would work a lot better with a picture, but hopefully it is understandable as it is.
The pieces can be made to be connected if you want too. To connect the bricks of a given colour, just bore holes through to the nearest neighbours of the same colour being careful that the holes avoid the 3-fold intersections - you can even ensure that each hole passes through just one brick of a different colour - and fill each bore hole with the same colour as the bricks that it connects between. Do this for each of the four colours, where of course you also need to avoid the (thin) boreholes that you have already made for one of the other colours.
In this case the pieces and some of their intersections have non-trivial first homology and second homology, and this is what keeps the cover from being too similar to its nerve.