One way to do this (though this might be the iterative method you reject) is to use the recurrence:
T[n+1]'/(n+1) - T[n-1]'/(n-1) = 2T[n] n>=2
This requires one to be able to compute the derivatives of the first 3 polynomials by hand, but since
T[0](x) = 1
T[1](x) = x
T[2](x) = 2*x*x-1
this is straightforward.
The coefficients in the recurrence are independent of x, so that if T[j,k] is the k'th derivative of the j'th Chebyshev poly, we can easily differentiate it, getting
T[n+1, k]/(n+1) - T[n-1,k]/(n-1) = 2T[n,k-1] n>=2
So the code could be:
compute the T[n,0] (ie the polynomials) at the point, for n=0..deg
initialise T[j,d] for j=0,1,2 and the required degrees
for j=1..deg
use the recurrence to compute the remaining polynomials