ASCII Dragon's Curve

Ruby, 239 201 bytes

This is a lambda function which should be called in the same manner as the one in the ungolfed version.

Golfing improvements include: assignment of 8<<n/2 to a variable for re-use; upto loop instead of each loop; ternary operator instead of if..else..end; use of [y,y+=d].max to calculate where to print the |; use of ?_ and ?| instead of the equivalent '|'and '_'; and elimination of redundant %4 (thanks Sp3000.)

->n{a=Array.new(m=8<<n/2){" "*m}
p=q=1+x=y=m/2
r=3
1.upto(1<<n){|i|d=(r&2)-1
r%2>0?(a[y][x+=d]=?_
x+=d):(a[[y,y+=d].max][x]=?|
p=x<p ?x:p
q=x>q ?x:q)
r+=i/(i&-i)}
a.delete(a[0])
puts a.map{|e|e[p..q]}}

It relies on the following formula from Wikipedia:

First, express n in the form k*(2^m) where k is an odd number. The direction of the nth turn is determined by k mod 4 i.e. the remainder left when k is divided by 4. If k mod 4 is 1 then the nth turn is R; if k mod 4 is 3 then the nth turn is L.

Wikipedia gives the following code:

There is a simple one line non-recursive method of implementing the above k mod 4 method of finding the turn direction in code. Treating turn n as a binary number, calculate the following boolean value: bool turn = (((n & −n) << 1) & n) != 0

I improved this to i/(i&-i)%4 which uses the same technique of using the expression i&-i to find the least significant digit but my expression gives 1 (for left turn)or 3 (for right turn) directly, which is handy as I track direction as a number 0..3 (in order north, west, south, east for golfing reasons.)

Ungolfed original in test program

f=->n{
  a=Array.new(8<<n/2){" "*(8<<n/2)}  #Make an array of strings of spaces of appropriate size 
  p=q=1+x=y=4<<n/2                   #set x&y to the middle of the array, p&q to the place where the underscore for n=0 will be printed.                             
  r=3                                #direction pointer, headed East
  (1..1<<n).each{|i|                 #all elements, starting at 1
    d=(r&2)-1                          #d is +1 for East and South, -1 for West and North
    if r%2>0                           #if horizontal
      a[y][x+=d]='_'                     #move cursor 1 position in direction d, print underscore,
      x+=d                               #and move again.
    else                               #else vertical
      a[(y+([d,0].max))][x]='|'          #draw | on the same line if d negative, line below if d positive
      y+=d                               #move cursor
      p=x<p ?x:p                         #update minimum and maximum x values for whitespace truncation later
      q=x>q ?x:q                         #(must be done for vertical bars, to avoid unnecesary space in n=0 case)
    end
    r=(r+i/(i&-i))%4                   #update direction
  }
  a.delete(a[0])                     #first line of a is blank. delete all blank lines.
  puts a.map!{|e|e[p..q]}                 #use p and q to truncate all strings to avoid unnecessary whitespace to left and right.
}


f.call(0)
f.call(2)
f.call(3)
f.call(11)

Python 2, 270 222 bytes

y=X=Y=0
i=m=x=1
D={}
k=2**input()
while~k+i:j=Y+(y>0);s={2*X+x};D[j]=D.get(j,s)|s;m=min(m,*s);Y+=y;X+=x;exec i/(i&-i)*"x,y=y,-x;";i+=1
for r in sorted(D):print"".join(" | _"[(n in D[r])+n%2*2]for n in range(m,max(D[r])+1))

Now using the formula for the nth turn. I saw the (((n & −n) << 1) & n) formula on Wikipedia, but didn't realise how useful it was until I saw it in @steveverrill's answer. I actually drop the %4 as well, so there's a lot of rotating going on, making larger inputs take a while.

---

Side remark: This isn't graphical output, but here's some golfed turtle code:

from turtle import*
for i in range(1,2**input()+1):fd(5);lt(i/(i&-i)*90)

C#, 337 bytes

There's a bit of rules abuse here. There's no restriction on leading space. Unfortunately, the canvas is finite, so there is an upper limit for n.

Indented for clarity:

using C=System.Console;
class P{
    static void Main(string[]a){
        int n=int.Parse(a[0]),d=2,x=250,y=500;
        var f="0D";
        while(n-->0)
            f=f.Replace("D","d3t03").Replace("T","10d1t").ToUpper();
        C.SetBufferSize(999,999);
        foreach(var c in f){
            n=c&7;
            d=(d+n)%4;
            if(n<1){
                var b=d%2<1;
                x+=n=b?1-d:0;
                y+=b?0:2-d;
                C.SetCursorPosition(x*2-n,y+d/3);
                C.Write(b?'_':'|');
            }
        }
    }
}