Truncate TreeForm to show only the top

Can use something like this:

ClearAll[showTopTree];
showTopTree[expr_, level_] :=
  Module[{myHold}, 
     SetAttributes[myHold, HoldAll];
     Function[code,
       TreeForm[Unevaluated@Unevaluated@code],
       HoldAll] @@
    (Hold[#] &@
      DeleteCases[MapAll[myHold, expr], _, {2*level, Infinity}] //.
           myHold[x__] :> x)];

Pretty ugly, but seems to work:

expr = Nest[1/(1 + #) (1 - #) &, w, 5]
Manipulate[showTopTree[expr, n], {n, 1, Depth[expr], 1}]

GraphicsGrid[Partition[showTopTree[expr, #] & /@ Range[6], 3]]

You can use the second argument of TreeForm to display and expression to a certain depth, so for your example you could do TreeForm[Nest[1/(1 + #) (1 - #) &, w, 5], 1] (although the result isn't very pretty in this case)

Edit

Instead of using TreeForm you could also construct a graph of the expression using ExpressionTreePlot in the GraphUtilities` package and use that to extract the desired subtree.

Needs["GraphUtilities`"];
exprTree[expr_] :=
 Module[{g, edges, labels},
  g = ExpressionTreePlot[expr, Top];
  edges = Rule @@@ Cases[g, Line[a_] :> a, Infinity][[1]];
  labels = Cases[g, Text[a_, b_] :> (b -> a[[1, 1]]), Infinity];
  {edges, labels}]

subTree[expr_, d_, pos_: Top] := Module[{edges, labels, sub},
  {edges, labels} = exprTree[expr];
  sub = NeighborhoodSubgraph[edges, 1, d];
  TreePlot[sub, pos, VertexRenderingFunction ->
    Function[{p, v}, 
     Text[Framed[Style[v /. labels, FontSize -> 10], 
       Background -> Lighter[Gray, .8]], p]]]]

Example:

subTree[Nest[1/(1 + #) (1 - #) &, w, 5], 4]

Here, I've chosen the style of VertexRenderingFunction in the definition of subTree to mimic the style of TreeForm but you could choose you own style for displaying the vertex labels.

The solutions seem a bit complicated. What about this one?

myTreeForm[expr_, dep_] := Map[Head, TreeForm[expr, dep], {dep + 1}];
a = Nest[1/(1 + #) (1 - #) &, w, 5];

myTreeForm[a, 1]
myTreeForm[a, 2]
myTreeForm[a, 3]