One way to approximate it is to get a best-fit ellipse by singular values decomposition. Since only the stretch factors are of interest we just compute the singular values. If I recall correctly it is their square roots that give the stretch. The method below requires that we center at the origin, hence the subtracting of the mean.

dataCentered = Map[# - Mean[data1] &, data1];
Sqrt[SingularValueList[dataCentered]]

(* Out[34]= {4.5038878147, 4.19597091008, 4.18475712383} *)

To check this we normalize and see if we get something that looks like the unit sphere.

dataNormed = Map[#/Sqrt[svals] &, dataCentered];
ListPointPlot3D[dataNormed, BoxRatios -> {1, 1, 1}]

I'm not sure what BoundedRegion does internally but it has a "FastEllipse" form specification. This is not a 'fit' as Daniel does (right?). i.e. with a fit you have data that is above/below the fit (or inside/outside) but this does not do that, it is always 'outside' of the data.

Depending what you are after this might be a alternative/better solution.

And an improved visualization:

lines = data1 /. {x_, y_, z_} -> Line[{{xc, yc, zc}, {x, y, z}}] /. 
   fit;

p2 = Show[plot, Graphics3D[lines]]

I find your way is easiest to follow as I am only looking for an approximate length. I am trying to figure out now how to find the diameter of the other dimension of the ellipsoid.

Hi I know this is not an efficient way to do it but I did this way. Note that data1=[{}]; is wrong notation, I changed to data1={{}};

 In[72]:= data3 = 
      Table[EuclideanDistance[data1[[j]], data1[[i]]], {i, 
        Length@data1}, {j, Length@data1}];
In[73]:= Max@data3

Out[73]= 8.50821

In[74]:= Position[data3, Max@data3]

Out[74]= {{5, 68}, {68, 5}}

In[75]:= data1[[5]]

Out[75]= {-1.47388, 2.66873, -0.915369}

In[76]:= data1[[68]]

Out[76]= {5.29808, -1.95733, 1.34955}



Show[HighlightMesh[ConvexHullMesh[data1], Style[2, Opacity[0.5]]], 
 Graphics3D[{Red, Line[{data1[[5]], data1[[68]]}]}], 
 Graphics3D[Point[data1]]]
Show[ListPointPlot3D[data1], 
 Graphics3D[{Red, Line[{data1[[5]], data1[[68]]}]}]]