[This may be largely an alternate version of Noam's answer, but the extra context could be
interesting.]

Let $N$ be the $m\times m$ matrix with $N_{i,i+1}=1$ for $i=1,\ldots,m-1$
and all other entries zero. Then the matrix
$$
A = \begin{pmatrix}0&I+N\\\\ (I+N)^T&0 \end{pmatrix}
$$
is the adjacency matrix of the path on $2m$ vertices. Now
$$
A^{-1} = \begin{pmatrix}
0&(I+N)^{-1}\\\\ (I+N)^{-T}&0
\end{pmatrix}
$$
and since $N^n=0$,
$$
(I+N)^{-1} = I-N+\cdots+(-1)^{n-1}N^{n-1}
$$
Let $D$ be the $2m\times 2m$ diagonal matrix with $D_{i,i}=(-1)^{i-1}$. Then it
easy to check that
$$
D^{-1}AD = \begin{pmatrix}
0&M\\\\ M^T&0
\end{pmatrix}
$$
where $M$ is the matrix from the question. The 2-norm we want is the
square of the largest eigenvalue of $D^{-1}AD$, which is the square of
the largest eigenvalue of $A$, which is the square of the reciprocal of the $n$-th
eigenvalue of the path on $2n$ vertices (which is its smallest positive eigenvalue).

The eigenvalues of the path on $n$ vertices are $2\cos\left(\frac{j\pi}{n+1}\right)$ for $j=1,\ldots,n$.

More on this appears in my old paper ``Inverses of trees''. (We can view $M$ as the incidence
matrix of a chain, and so some of the above extends to a larger class of posets.)