From my understanding of what you are asking, you may take the product of two depolarization operations, using the reduced density matrix of each qubit in the Bell state in the expression.
Let's denote our two qubits as $\mathrm{A}$ and $\mathrm{B}$. The Bell state of these two qubits is then:
$$
|\beta_{00} \rangle =\frac{|0 \rangle_\mathrm{A} \otimes |0 \rangle_\mathrm{B} + |1 \rangle_\mathrm{A} \otimes |1 \rangle_\mathrm{B}}{\sqrt{2}} = \frac{|00 \rangle + |11 \rangle}{\sqrt{2}}
$$
With a density matrix:
$$
\rho = | \beta_{00} \rangle \langle \beta_{00} | = \frac{ |00 \rangle \langle 00| + |00 \rangle \langle 11 | + |11 \rangle \langle 00| + |11 \rangle \langle 11 |}{2}
$$
Which makes the two reduced density matrices:
$$
\rho_\mathrm{A} = \text{tr}_\mathrm{B}(\rho) = \frac{|0 \rangle \langle 0| + |1 \rangle \langle 1|}{2} = \frac{I}{2}
$$
$$
\rho_\mathrm{B} = \text{tr}_\mathrm{A}(\rho) = \frac{|0 \rangle \langle 0| + |1 \rangle \langle 1|}{2} = \frac{I}{2}
$$
The operation can then be defined as (assuming the same noise parameter $Q$ on both applications of the operation):
$$
\mathcal{E} \otimes \mathcal{E} = (Q \; \frac{I}{2} + (1-Q) \rho_\mathrm{A}) \otimes (Q \; \frac{I}{2} + (1-Q) \rho_\mathrm{B})
$$
Which ultimately simplifies to:
$$
\mathcal{E} \otimes \mathcal{E} = Q(2-Q) \; \frac{I}{2} \otimes \frac{I}{2} + (1-Q)^2 \; \rho
$$
Because both reduced density matrices for the Bell state given are equivalent to the density matrix for a completely mixed state, i.e. $\frac{I}{2}$.
Hope this helps!