Here is a revised solution, for...
... n =  17  31
or, using letters to represent disks,
abcdefghijklmnopqrstuvwxyzABCDE
, ...
...which (again) seems like the maximum to me. 
has been
verified by Molhan
as being maximal.
Trivial steps have been condensed.
Step 1 2 3 4 5 6
abcdefghijklmnopqrstuvwxyzABCDE | | | | |
1 efghijklmnopqrstuvwxyzABCDE | | abcd | |
2 ijklmnopqrstuvwxyzABCDE | efgh abcd | |
3 ijklmnopqrstuvwxyzABCDE | abcdefgh | | |
4 mnopqrstuvwxyzABCDE | abcdefgh ijkl | |
5 opqrstuvwxyzABCDE | abcdefgh ijkl mn |
6 qrstuvwxyzABCDE op abcdefgh ijkl mn |
7 qrstuvwxyzABCDE mnop abcdefgh ijkl | |
8 qrstuvwxyzABCDE mnop abcdefgh kl ij |
9 qrstuvwxyzABCDE klmnop abcdefgh | ij |
10 qrstuvwxyzABCDE ijklmnop abcdefgh | | |
11 qrstuvwxyzABCDE ijklmnop cdefgh | ab |
12 qrstuvwxyzABCDE ijklmnop efgh cd ab |
13 qrstuvwxyzABCDE ijklmnop efgh abcd | |
14 qrstuvwxyzABCDE ijklmnop gh abcd ef |
15 qrstuvwxyzABCDE ghijklmnop | abcd ef |
16 qrstuvwxyzABCDE efghijklmnop | abcd | |
17 qrstuvwxyzABCDE abcdefghijklmnop | | | |
18 uvwxyzABCDE abcdefghijklmnop | qrst | |
19 wxyzABCDE abcdefghijklmnop | qrst uv |
20 yzABCDE abcdefghijklmnop wx qrst uv |
21 yzABCDE abcdefghijklmnop uvwx qrst | |
22 yzABCDE abcdefghijklmnop uvwx st qr |
23 yzABCDE abcdefghijklmnop stuvwx | qr |
24 yzABCDE abcdefghijklmnop qrstuvwx | | |
25 ABCDE abcdefghijklmnop qrstuvwx | yz |
26 CDE abcdefghijklmnop qrstuvwx AB yz |
27 CDE abcdefghijklmnop qrstuvwx yzAB | |
28 E abcdefghijklmnop qrstuvwx yzAB CD |
29 | abcdefghijklmnop qrstuvwx yzAB CD E
These steps may be reversed,
exchanging the roles of pegs 1 and 6,
to complete moving the whole tower from peg 1 to peg 6.
This approach was derived by mulling over
what happens after the symmetric midpoint,
step 29 above. At that stage...
... the last disk, E
, moves to a new peg, leaving peg 1 empty.
No other pegs should be empty, or else we didn't find the maximum n
because we would have the opportunity to move another disk
that could have been under E
.
So, how tall a stack can be moved atop E
when one peg is free?
After moving that stack, two pegs will be free,
as the bottom of the moved stack
would have been the D
that winds up resting on E
.
This, in turn, raises the question of
how many disks could subsequently be moved
onto the increasing goal tower above E
?
This is repeated to reveal the following recursive pattern.
| - | - | - - - - - pegs - - - - - - - - |
1 2-tall Step +0. | CD (..........other occupied pegs..........) E
free stack Step +1. C D (..........other occupied pegs..........) E
peg may be Step +2. | | (..........other occupied pegs..........) CDE
moved
2 4-tall Step +0. | | yzAB (.........occupied pegs.........) CDE
free stack Step +1. | yz AB (.........occupied pegs.........) CDE
pegs may be . . . Seen in reverse, step +1 was just .
moved . . . like asking how tall a stack can .
. . . go onto AB when 1 peg is empty. .
. . . Answer = 2 tall (yz). .
. . . And now we have an instance of .
. . . 1 free peg, which allows the .
. . . 2-tall stack AB to go atop CDE. .
Step +2. | yz | (.........occupied pegs.........) ABCDE
Step +3. | | | (.........occupied pegs.........) yzABCDE
3 8-tall Step +0. | | | qrstuvwx (..occupied pegs..) yzABCDE
free stack Step +1. | | qrst uvwx (..occupied pegs..) yzABCDE
pegs may be . . . . Seen in reverse, step +1 .
moved . . . . was just like asking how tall .
. . . . a stack can go on uvwx when .
. . . . 2 pegs are empty. Answer = .
. . . . 4 tall (qrst). Similar .
. . . . recursion on 2 free pegs .
. . . . puts uvwx on yzABCDE. .
Step +2. | | qrst | (..occupied pegs..) uvwxyzABCDE
Step +3. | | | | (..occupied pegs..) qrstuvwxyzABCDE
4 16-tall This 16 is derived similarly, completing the total 1+2+4+8+16 = 31
free stack may disks because the last stack-move begins with 4 empty pegs and ends
pegs be moved ends with 5 empty pegs along with the moved tower on the 6th peg.
Sketch of a proof that this is optimal,
thanks to Mike Earnest:
Let F(k) be size of the largest stack
that can be moved off an infinitely tall stack
when k empty pegs are available,
which makes
F(3) = 8
similar, in the method chart above, to
an "8-tall stack may be moved"
with "3 free pegs."
It can be shown that
F(k) ≤ 1 + F(1) + F(2) + ... + F(k−1)
by considering what pegs look like at midpoints of the process.