I found 125843706679.

For any 10-digit pandigital, d(n), s(n), c(n) will always be [45, 285, 2025], but this can't be a 10 digit pandigital due to multiple appearances of several digits

For an 11-digit pandigital, there are 10 possibilities for each function:

[45, 285, 2025] if 0 is the digit that appears twice

[46, 286, 2026] if 1 is the digit that appears twice

[47, 289, 2033] if 2 is the digit that appears twice

[48, 294, 2052] etc

[49, 301, 2089] ...

[50, 310, 2150]

[51, 321, 2241]

[52, 334, 2368]

[53, 349, 2537]

[54, 366, 2754] if 9 is the digit that appears twice

But a quick review of those 10 triplets shows that each has too many of some digit.

For a 12-digit pandigital, there are 55 options for the 2 extra digits.

The program below produces a list of each of those, identified by the 2 extra digits and the triplet of sum of digits, squares, cubes of digits.

I had not yet worked out an algorithm for checking these for inconsistencies, so I scanned the 55 visually and found only one that works:

extra digits; [SOD, SOS, SOC]

[[6, 7], [58, 370, 2584]]

The sum of digits is included in the sum of cubes of digits.

2584

370

Digits not yet used:  16679 

My Solution:  125843706679

The program below will accept any number of digits, as the digits go from 11, 12, 13 etc, the possibilities go from:  10, 55, 220, 715, 2002   So if I did not find a 12-digit pandigital, I would have needed to come up with an algorithm to weed out those that have superfluous digits.  Weeding out some would be easy, but non-trivial to weed out all.

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a = 0

b = 0

c = 0

trip10 = [0,0,0]

for i in range(10):

    trip10[0] += i

    trip10[1] += i**2

    trip10[2] += i**3

sods = []

soss = []

socs = []

triplets = []

digits = '0123456789'

digs = [int(a) for a in digits]

from itertools import combinations_with_replacement

ndigits = 12

extra = ndigits - 10

allints = []

for comb in combinations_with_replacement(digits, extra):

    ints = [int(c) for c in comb ]

    allints.append(ints)

    trip = trip10[:]

    for i in ints:

        trip[0] += i

        trip[1] += i**2

        trip[2] += i**3

    triplets.append([ints,trip])

for t in triplets:

    print(t)

-------   triplets for n = 12