TL;DR: The threshold is not just a property of a quantum error correcting code. The existence and value of the threshold depends on the noise model (and also on the chosen decoding procedure). To demonstrate this, I exhibit a noise channel under which the no-cloning theorem implies that $p_T<\frac12$ for every code and every decoder. This rules out the existence of a code with $p_T=1$ for every noise model.
Digression: break-even vs threshold
Before describing the noise process that yields $p_T<\frac12$, let me point out that the condition $p=p_L$ does not define the threshold. Instead, that condition defines the break-even point, which is not necessarily the same.
In the simplest terms, the threshold is whatever number appears in the relevant instance of the threshold theorem. More helpfully, the threshold is the largest real number $p_T\in[0,1]$ such that a quantum computer whose physical gate operations fail with probability $p<p_T$ can achieve arbitrarily low probability $\epsilon$ of failure in executing any quantum program with $K$ operations at the expense of overhead that is $\mathcal{O}\left(\mathrm{polylog}^c\left(\frac{K}{\epsilon}\right)\right)$ per operation.
Note the asymptotic character of the threshold theorem. It means that in practice looking for the threshold involves a (more or less explicit) computation of the limit of a series. In more practical terms, suppose you draw the $p_L$ vs $p$ curves for a family of quantum error correcting codes with increasing code distance. Each of the curves will typically intersect the line $p_L=p$ at some point. This is the break-even point for that particular code distance. The lines will also intersect each other and these intersection points may converge. If they do, then typically they converge from the right. The $p$ coordinate of the limit point is the threshold, because for any $p$ that is less than that value we can guarantee that a modestly larger quantum computer will have a substantially lower logical error rate than a smaller quantum computer. There are codes for which the limit point lies at $p=0$. We say that those codes don't have a threshold. Nevertheless, these codes may have a positive break-even point at every code distance.
I now return to the main point of constructing an error model under which no code can achieve $p_T
\geqslant\frac12$.
Intuitive idea
The key idea is that noise, which we usually think of as undesired deviations of quantum dynamics from theoretical ideal due to system imperfections, can in fact be due to the environment secretly "spying" on our encoded quantum information. If we were successful in our attempts to preserve a quantum state in the code subspace and if the environment were simultaneously successful in its attempts to spy on our quantum state, then jointly we and the environment would have copied the state!
Probabilistic $\text{SWAP}$ noise
Consider a two-qutrit quantum channel $\mathcal{E}$ which extends the action of
$$
\mathcal{E}(\rho\otimes|2\rangle\langle 2|)=\frac{\rho\otimes|2\rangle\langle 2|}{2}+\frac{|2\rangle\langle 2|\otimes\rho}{2}.\tag1
$$
We can easily realize this channel by swapping the qutrits when a symmetric coin comes up heads and doing nothing if it comes up tails. Suppose we use the $|0\rangle$ and $|1\rangle$ states of each qutrit to encode a qubit. Then the error channel $\mathcal{E}$ inflicts an erasure error on the first qubit with probability $p=\frac12$.
We can also consider a virtual qubit that begins on the left side at the input of $\mathcal{E}$ and migrates to the right at the output of $\mathcal{E}$. The channel $\mathcal{E}$ inflicts an erasure error on this virtual qubit with probability $p=\frac12$, too.
Arbitrarily reliable universal copier from encoding and recovery
Suppose we have two blocks of $n$ qubits where each qubit resides in the $\mathrm{span}(|0\rangle, |1\rangle)$ subspace of a qutrit. Suppose further that we use a quantum error correcting code $\mathcal{C}$ to encode one logical qubit into each block. Let $\mathcal{R}$ denote the code's recovery operation. Suppose that the joint effect of noise on each of the $n$ corresponding pairs of physical qutrits in the two logical code blocks is described by $\mathcal{E}$. Finally, assume that $\mathcal{R}$ achieves threshold $p_T\geqslant\frac12$ under the noise described by $\mathcal{E}$.
Let's initialize the two code blocks in $\rho\otimes|2\rangle\langle 2|^{\otimes n}$ for some logical state $\rho$. We will attempt recovery for two logical qubits: the one encoded in the left code block and the virtual one encoded in the $n$ virtual qubits that begin on the left at the input to $\mathcal{E}$ and exit on the right at its output.
Note that the left logical qubit and the virtual logical qubit are both seeing physical erasure errors occurring with probability $p=\frac12$, so noise and recovery yield
$$
\left[(\mathcal{R}\otimes\mathcal{R})\circ\mathcal{E}^{\otimes n}\right](\rho\otimes|2\rangle\langle 2|^{\otimes n})=(1-2p_L+p_L^2)\rho\otimes\rho+(2p_L-p_L^2)\sigma\tag2
$$
where the first term corresponds to successful decoding of both logical qubits and $\sigma$ is an error term. Note that the logical error probability $p_L\lt p=\frac12\leqslant p_T$. Since this is below the threshold for each of the logical qubits, we can use concatenation $\mathcal{C}$ (or simply a larger code family member if $\mathcal{C}$ is a topological code) to construct a channel
$$
\left[(\mathcal{R}^{(k)}\otimes\mathcal{R}^{(k)})\circ\mathcal{E}\right](\rho\otimes|2\rangle\langle 2|^{\otimes n})=(1-\epsilon)\rho\otimes\rho+\epsilon\sigma\tag3
$$
where $\mathcal{R}^{(k)}$ denotes the recovery operation of $k$ levels of concatenation of the code $\mathcal{C}$, which has arbitrarily low error rate $\epsilon$. Clearly, this channel copies an arbitrary quantum state with arbitrarily high reliability and is therefore impossible. Therefore, the assumption that $p_T\geqslant\frac12$ is false.
