An example which disproves $\|p(T)\|_{op} = \|p\|_\sup$ in arbitrary Banach spaces (Gelfand theory). Let me go straight to the question and then write some discussion about it.

Question: Does there exist a Banach space $X$, an operator $T:X\rightarrow X$ with operator norm $\|T\|_{op}\leq 1$ and a sequence of polynomials $p_n:\mathbb{C}\rightarrow \mathbb{C}$ so that $\sup_{x\in D} |p_n(x)|\rightarrow 0$ as $n$ goes to infinity, where $D$ is the unit dist, but $\|p_n(T)\|_{op}=1$ (or any constant greater than zero that is independent on $n$)? Equivalently, the map $p\mapsto p(T)$ from $(\mathrm{Poly}(D),\|\cdot\|_\infty)$  to $(\mathcal{L}(X),\|\cdot\|_{op})$ is not continuous.

Two notes:
(1) It is not likely, but it could be that there are no such examples, in that case a proof or reference would be nice :-)
(2) It is much appreciated if you can provide an example in the case where $X=L^p(\Omega)$ for every $p>2$, where $\Omega$ is some measured space (say $\Omega=[0,1]$).
Discussion:
I do know that there is no counterexample in the case where $X$ is a Hilbert space (edit: and $T$ is normal), but that result requires the use of Gelfand theory, which is not valid for arbitrary Banach spaces (because we do not have natural $C^\star$ algebras).
Let $H$ be a Hilbert space and $T:H\rightarrow H$ be a continuous operator. For a polynomial $p(x) = \sum_n a_n x^n$ we can define a new operator $p(T) = \sum_n a_n\cdot T^n$ on $H$.
A well known, seemingly trivial, but extremely deep result in functional analysis is the fact that the operator norm of $p(T)$ is exactly the supremum of $p$ on the spectrum of $T$.
The main key point in the proof is the fact that for an operator $T$, if $\sigma(T)$ is the spectrum of $T$ then $$\|T\|_{op} = \max \sigma(T).$$ This combined with the fact that $\sigma(p(T))=p(\sigma(T))$ gives the assertion above immediately. (The equality, $\|T\|_{op} = \max \sigma(T)$, does not seem to be true for arbitrary Banach spaces, at least I fail to incorporate the original proof, because Gelfand theory is not valid.)
Now, if we assume that $\|T\|_{op}\leq 1$, we can already deduce that $\sigma(T)\subseteq D$. Therefore, if $p_n$ are polynomials so that $\|p_n\|_{\infty,D}\rightarrow 0$, then $\|p_n(T)\|_{op}\leq \|p_n\|_{\infty,D}\rightarrow 0$. This proves that if $X$ is an Hilbert space then there is no example of the form I am looking for.
I suspect that this is not true for arbitrary Banach spaces. A concrete example will help me greatly.
 A: The statement that "the operator norm of $p(T)$ is exactly the supremum of $p$ on the spectrum of $T$" is not always
true, as it requires the operator to be normal.  Nevertheless it is indeed true that, on Hilbert's space, one has that
$$
  \|p(T)\|\leq \sup_{|z|\leq 1}|p(z)|, \tag{1}
  $$
for any polynomial $p$, and every contraction  $T$ (meaning an operator such that $\|T\|\leq 1$).  This is actually known as
von Neumann's inequality.
In case $T$ is normal, the above can be proved by Gelfand's theory and spectral radius, but the technique that works for
arbitrary contractions is
based on unitary dilation.
For general Banach spaces (1) may indeed fail very badly, in the sense that there are contractions $T$ for which
there is no positive constant
$C$ satisfying
$$
  \|p(T)\|\leq C\sup_{|z|\leq 1}|p(z)|, \tag{2}
  $$
for every polynomial $p$.  The fact that no such constant exists is clearly equivalent to
saying that there exists a sequence $\{p_n\}_n$ of polynomials such that $\sup_{|z|\leq 1}|p_n(z)|\to 0$, and
$\|p_n(T)\|=1$.
Here is an example: let $S$ be the shift operator on $\ell ^1$, namely the map given by
$$
  S(x_1, x_2, x_3, \ldots ) = (0, x_1, x_2, \ldots ),\quad\forall (x_1, x_2, x_3, \ldots ) \in \ell ^1.
  $$
Given a polynomial $p$, say
$p(z)=\sum_{k=0}^na_kz^k$,
notice that
$$
  \|p(S)\|\geq
  \|p(S)(1,0,0,\ldots )\|= \|(a_0,a_1,\ldots ,a_n,0,\ldots )\|=\sum_{k=0}^n|a_k|.
  $$
Thus, assuming by contradiction that (2) is true for some $C$, we deduce that
$$
  \sum_{k=0}^n|a_k|\leq C\sup_{|z|\leq 1}|p(z)|, \tag{3}
  $$
for every polynomial $p$, and we will next show that this can't happen.
The rest of this answer is based on many highly relevant comments made by a variety of users after an initial and rather
insipid atempt at proving the impossibility of (3).
In order to prove that no constant $C$ exists satisfying (3), we will provide a sequence $\{p_n\}_n$ of polynomials, such
that the right-hand-side of (3) is uniformly bounded, while the left-hand-side blows up.
For each natural number $n\geq 1$,  consider the polynomial
$$
  p_n(z) = \sum_{k\in [0, 2n]\setminus\{n\}}\frac{z^k}{k-n},
  $$
where we observe that the exclusion of the summand corresponding to $k=n$ is necessary to avoid a division by zero.
I should also say that the above polynomials were obtained by a process of reverse engineering based on a suggestion by
@RyszardSzwarc.  Thanks Ryszard!
We then claim that
$$
  \sup_{n\in {\mathbb N}}\sup_{|z|\leq 1}|p_n(z)| = \sup_{n\in {\mathbb N}}\sup_{|z|=1}|p_n(z)|<\infty . \tag{4}
  $$
The first equality is an easy consequence of the maximum modulus principle, while the inequality on the right is proven as follows:
for every $n$, and every $z$, with $|z|=1$,  we have that
$$
  z^{-n}p_n(z) =
  \sum_{k\in [0, 2n]\setminus\{n\}}\frac{z^{k-n}}{k-n} =
  \sum_{k\in [-n, n]\setminus\{0\}}\frac{z^{k}}{k} = $$$$ =
  \sum_{k=1}^n\frac{z^{-k}}{-k} +   \sum_{k=1}^n\frac{z^{k}}{k} =
  \sum_{k=1}^n\frac{z^{k}-z^{-k}}k.
  $$
Writing   $z=e^{it}$, for some real number $t$,  we deduce that
$$
  |p_n(z)| =  |z^{-n}p_n(z)| =
  \Big |\sum_{k=1}^n\frac{e^{ikt}-e^{-ikt}}k\Big | =
  \Big |\sum_{k=1}^n\frac{\sin(kt)}k\Big |,
  $$
which is uniformly bounded (independently of $n$) by this answer, hence proving claim (4).
On the other hand, the left-hand-side of (3), namely the
sum of the absolute values of the coefficients of $p_n$, is given by
$$
  \sum_{k\in [0, 2n]\setminus\{n\}}\frac 1{|k-n|} = 2\sum_{k=1}^n\frac{1}{k},
  $$
which goes to infinity, as $n\to \infty $.  This proves that there cannot exist a constant $C$, as in (3).
