= Solution
The <centered quadratic spline interpolation> <matrix> is the <symmetric tridiagonal Toeplitz matrix>
$$
\boxed{A_x=\frac18\begin{pmatrix}
6&1&0&\cdots&0\\
1&6&1&\ddots&\vdots\\
0&1&6&\ddots&0\\
\vdots&\ddots&\ddots&\ddots&1\\
0&\cdots&0&1&6
\end{pmatrix}.}
$$
For $n=1$, this means simply $A_x=(3/4)$. To evaluate the inverse <norm> exactly, let $D=\operatorname{diag}((-1)^i)_{i=1}^n$ and $M=8DA_xD=\operatorname{tridiag}(-1,6,-1)$. If $J$ is the nonnegative adjacency <matrix> of the path, then $M=6I-J$ and $\|J/6\|_\infty\le1/3$. The convergent <Neumann series>
$$
M^{-1}=\frac16\sum_{r=0}^{\infty}(J/6)^r
$$
therefore has nonnegative entries. Since $A_x^{-1}=8DM^{-1}D$, its entrywise absolute values are $8M^{-1}$. Let $r_i=\sum_j|(A_x^{-1})_{ij}|$. Then $r=8M^{-1}\mathbf1$, or
$$
6r_i-r_{i-1}-r_{i+1}=8,\qquad r_0=r_{n+1}=0.
$$
The homogeneous characteristic roots are $3\pm2\sqrt2$. Put $\tau=3-2\sqrt2\in(0,1)$. The particular solution $r_i=2$ and the two boundary conditions give
$$
r_i=2\left(1-\frac{\tau^i+\tau^{n+1-i}}{1+\tau^{n+1}}\right).
$$
These row sums are symmetric and increase toward the centre: the exponential sum in the numerator is minimized at the middle <integer> or pair of <integers>. With $m=\lfloor(n+1)/2\rfloor$, the exact finite-size inverse <norm> is thus
$$
\boxed{\|A_x^{-1}\|_{\ell^\infty}
=2\left(1-\frac{\tau^m+\tau^{n+1-m}}{1+\tau^{n+1}}\right)<2.}
$$
For example it equals $4/3$ at $n=1$ and $8/5$ at $n=2$, tending to $2$ as $n\to\infty$. Stating only the bound two would not evaluate the finite <matrix> <norm> requested here.
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