= Solution
Order the <interpolation nodes> as $t_0<\cdots<t_n$, as required by their alternating nodal data. Write $w(x)=\prod_{j=0}^n(x-t_j)$ and $\ell_i(x)=w(x)/[(x-t_i)w'(t_i)]$ for the <Lagrange cardinal polynomials>. Differentiating <Lagrange interpolation> gives
$$
p^{(k)}(x)=\sum_{i=0}^np(t_i)\ell_i^{(k)}(x),\qquad M_k(x)=\sum_{i=0}^n|\ell_i^{(k)}(x)|.
$$
The upper bound is the <triangle inequality>; equality follows by choosing each independent nodal value to be the sign of its weight. Thus this is the <nodal derivative norm from Lagrange cardinal polynomials>.
To identify the maximizing signs, put $P_i=w/(x-t_i)$, a <monic polynomial> of degree $n$. Since $\operatorname{sign}w'(t_i)=(-1)^{n-i}$,
$$
(-1)^i\ell_i^{(k)}(x)=\frac{(-1)^nP_i^{(k)}(x)}{|w'(t_i)|}.
$$
For $0\le k<n$, let $\xi_{i,j}$ be the ordered roots of $P_i^{(k)}$, $1\le j\le n-k$. The roots of $P_i$ decrease componentwise as $i$ increases: deleting a later node replaces one retained node by an earlier one. The <monotonicity of polynomial critical points in their roots>, iterated $k$ times, therefore gives $\xi_{n,j}\le\cdots\le\xi_{0,j}$. Moreover, $P_n$ and $P_0$ weakly interlace, so the <Markov interlacing lemma> gives $\xi_{0,j}\le\xi_{n,j+1}$. Together,
$$
\xi_{n,j}\le\xi_{n-1,j}\le\cdots\le\xi_{0,j}\le\xi_{n,j+1}.
$$
These root bands have disjoint interiors. At a fixed $x$, all but at most one band lie wholly on one side of $x$. In the possible remaining band, $\xi_{i,j}$ moves monotonically as $i$ increases. Since the <leading coefficients> of all $P_i^{(k)}$ are positive, the sequence $P_i^{(k)}(x)$ has at most one sign change after zero entries are ignored. The same holds for $(-1)^i\ell_i^{(k)}(x)$. If $k=n$, every $P_i^{(n)}=n!>0$, so the conclusion also holds.
The maximizing nodal signs can consequently be chosen as $\pm(-1)^i$ throughout, or as $\pm(-1)^i$ before one index $s$ and $\mp(-1)^i$ from that index onward. Zero weights can be assigned either sign to complete such a pattern. These are precisely the nodal values of $\pm p^*$ or $\pm q_s$, proving
$$
\boxed{M_k(x)=|p^{*\,(k)}(x)|\quad\text{or}\quad M_k(x)=|q_s^{(k)}(x)|\text{ for some }1\le s\le n}.
$$
The argument works for all real $x$, hence in particular for $[-1,1]$. For $k>n$ every derivative vanishes and the result is immediate.
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