As pointed out in the comments, you cannot use the one-qubit formula because something like $X\rho X$ does not make sense if $\rho$ is a 2-qubit state. In fact, for this reason the answer you based your question on is wrong, as well. While the other answer on that old post is correct (up to the typo pointed out in a comment), it does not give any derivation for that decomposition. I'd like to fill that gap here.
The key is the following identity which, to my knowledge, was first proven in the 1976 paper by Gorini, Kossakowski, and Sudarshan, Equation 2.5 therein:
Lemma. Given any orthonormal basis $\{G_k\}_k$ of $\mathbb C^{n\times n}$ (that is, a basis which satisfies ${\rm tr}(G_j^\dagger G_k)=\delta_{jk}$ for all $j,k$) it holds that
$$
\sum_k G_k\rho G_k^\dagger={\rm tr}(\rho)I_n
$$
for all $\rho\in\mathbb C^{n\times n}$.
Therefore, to obtain the operator-sum representation of your channel all you need is an orthonormal basis for 2-qubits (i.e., of $\mathbb C^{4\times 4}$). One such choice would be to extend the 1-qubit Pauli basis $\{\frac1{\sqrt2}I_2,\frac1{\sqrt2}X,\frac1{\sqrt2}Y,\frac1{\sqrt2}Z\}$—after re-writing $P_0:=I_2,P_1:=X,P_2:=Y,P_3:=Z$—to the 2-qubit Pauli basis
$$
\Big\{\frac1{\sqrt2}P_j\otimes\frac1{\sqrt2}P_k\Big\}_{j,k=0}^3=\Big\{\frac1{2}P_j\otimes P_k\Big\}_{j,k=0}^3\,.
$$
As such, your $\mathcal E$ can be written as
\begin{align*}
\mathcal{E}(\rho) &= (1-\lambda)\rho + {\rm tr}(\rho)\frac{\lambda I_4}{4}\\
&=(1-\lambda)\rho+\frac\lambda4\sum_{j,k=0}^3\Big(\frac{P_j\otimes P_k}2\Big)\rho\Big(\frac{P_j\otimes P_k}2\Big)^\dagger\\
&=\Big(1-\frac{15\lambda}{16}\Big)\rho+\frac\lambda{16}\sum_{\substack{j,k=0\\(j,k)\neq (0,0)}}^3(P_j\otimes P_k)\rho(P_j\otimes P_k)^\dagger\\
&=\Big(\sqrt{1-\frac{15\lambda}{16}}I_4\Big)\rho\Big(\sqrt{1-\frac{15\lambda}{16}}I_4\Big)^\dagger+\sum_{\substack{j,k=0\\(j,k)\neq (0,0)}}^3\Big(\frac{\sqrt\lambda}4P_j\otimes P_k\Big)\rho\Big(\frac{\sqrt\lambda}4P_j\otimes P_k\Big)^\dagger.
\end{align*}
Of course, because all Paulis are Hermitian we can just drop the ${}^\dagger$ in this case.
As a final remark, there was nothing particular about the Pauli basis: because our key lemma works for any basis we could also have used the computational basis $\{|i\rangle\langle j|\}_{i,j}$, cf. this answer.
