Edited:
@Turbo: You're right. I was assuming that the two codewords must be nonzero. In general, when one codeword is zero, one gets a weaker lower bound of the form
$N_d {q^k \choose T}$
and the rest of the argument follows with a linear $N_d$ instead of a quadratic factor $N_d^2/2.$
A special case (original answer):
In the specific case of a binary linear code $C$ the nonzero values of the weight distribution coefficients $A_w=\#\{c \in C:wt(c)=w \}$ are $A_0=1,A_d=N_d,\ldots$. Now assume the code contains the all $1$ vector as a codeword. Then $A_{n-d}$ is also nonzero and $A_{n-d}\geq A_d=N_d$ since the sums (or differences) $(1,1,\ldots,1)+c$ are also codewords in $C.$
Thus there are at least $N_d(N_d-1)/2$ pairs $c_i,c_j$ in $C$ which satisfy $|c_i-c_j|=d$.
For each one of these pairs, you can select the remaining $T-2$ codewords (assumed distinct) arbitrarily provided $T-2 \leq 2^k.$ So you have approximately
$$ (N_d^2/2) {2^k \choose T} \qquad (1)$$
which should be at least $O(2^k)$ provided $T$ is small enough compared to $2^k.$ Of course, if the code is also cyclic, then $N_d \geq n$ and this helps extend the validity regime of the lower bound. i seem to recall there are some sporadic linear codes with $N_d=3,$ and $n$ long, but can't think of the reference right now.
If you're interested in asymptotics, quite a few linear codes have weight distribution that is "asymptotically normal". This means that
$$A_i/2^k \approx 2^{-n} {n \choose d}$$
and this should help refine the estimate in (1) above when you let $i=d$. See Chapter 9, especially section 10 of McWilliams and Sloane for more.
There should be obvious generalizations of this argument to nonbinary characteristic, such as assuming any Hamming weight $n$ codeword belongs to $C.$
