Let's see if this works.

So I think the answer to my question is positive.

Let's take an injective resolution $\mathcal{F}\rightarrow \mathcal{I}^\bullet$ of $\mathcal{F}$ an let $K_n^\bullet:=H^0_{Z_n}(X,\mathcal{I}^\bullet)$ be a complex which computes $H^*_{Z_n}(X,\mathcal{F})$. Then be $K_n^\bullet\rightarrow J_n^{\bullet,\bullet}$ a Cartan-Eilenberg resolution of $K_n^\bullet$, such that $K^p\rightarrow J^{p,\bullet}$ is an injective resolution. Then, by definition, $$H^*(\mathrm{Tot}(\varprojlim_{n}J_n^{\bullet,\bullet}))=H^*(R\varprojlim_nK_n^\bullet).$$

The double complex $\varprojlim_n J_n^{\bullet,\bullet}$ induces by the two standard filtrations the mentioned hypercohomology spectral sequences

\begin{align} {}^IE_1^{pq}&={\varprojlim_n}^{(q)}H^0_{Z_n}(X,\mathcal{I}^p) \Rightarrow H^{p+q}(R\varprojlim_nK_n^\bullet), \\ {}^{II}E_2^{pq}&={\varprojlim_n}^{(p)}H^q\big(H^0_{Z_n}(X,\mathcal{I}^\bullet)\big) \Rightarrow H^{p+q}(R\varprojlim_nK_n^\bullet). \end{align}

By the arguments from the Proof of Proposition 4 in Section 2 of https://ivv5hpp.uni-muenster.de/u/pschnei/publ/pap/xsymm.pdf, we know

$${}^IE_1^{pq}=0 \text{ for } q \neq 0, \,\,\,\,\,\,\,\,\,(1)$$ $$ {}^{II}E_2^{pq}=0 \text{ for } p \neq 0,1. \,\,\,(2)$$

Hence, both spectral sequences degenerate on the second page.

Additionally, let us fix $n \in \mathbb{N}$ and set

\begin{align} {}^IE_0^{'pq}&=J_n^{p,q} \Rightarrow H^{p+q}\big(\mathrm{Tot}(J_n^{\bullet,\bullet})\big)=H^{p+q}_{Z_n}(X,\mathcal{F}), \\ {}^{II}E_0^{'pq}&=J_n^{q,p} \Rightarrow H^{p+q}\big(\mathrm{Tot}(J_n^{\bullet,\bullet})\big)=H^{p+q}_{Z_n}(X,\mathcal{F}). \end{align}

Furthermore we have a natural morphism of double complexes

$$ \varprojlim_n J_n^{\bullet,\bullet} \rightarrow J_n^{\bullet,\bullet}.$$

As mentioned in EGA III, p. 30, a morphism of double complexes induces a natural morphism of spectral sequences (for both filtrations)

This implies we have a morphism $ {}^IE_1^{\bullet \bullet} \rightarrow {}^IE_1^{'\bullet \bullet}$ which is nothing but the morphism of complexes

$$ H^0_{Z}(X,\mathcal{I}^\bullet)={\varprojlim_n} H^0_{Z_n}(X,\mathcal{I}^\bullet) \hookrightarrow H^0_{Z_n}(X,\mathcal{I}^\bullet)$$ inducing the natural morphism $\varphi:H^*_Z(X,\mathcal{F})\rightarrow H^*_{Z_n}(X,\mathcal{F})$ in cohomology. The mentioned equality follows also by arguments from https://ivv5hpp.uni-muenster.de/u/pschnei/publ/pap/xsymm.pdf. Furthermore, the first spectral sequence implies $H^*_Z(X,\mathcal{F})=H^{*}(R\varprojlim_nK_n^\bullet).$

Then by the induced morphism on spectral sequences and the degeneration on the second page we have a commutative diagram

$\require{AMScd}$ \begin{CD} H^{n}(R\varprojlim_nK_n^\bullet) @>>> H^*_{Z_n}(X,\mathcal{F})\\ @V V V= @VV V\\ {}^{II}E_2^{0n} @>>> {}^{II}E_2^{'0n} \end{CD}

which is

$\require{AMScd}$ \begin{CD} H^*_Z(X,\mathcal{F})@>\varphi>> H^*_{Z_n}(X,\mathcal{F})\\ @V g V V= @VV id V\\ {\varprojlim_n} H^*_{Z_n}(X,\mathcal{F})@>p_n>> H^*_{Z_n}(X,\mathcal{F}) \end{CD} what we were looking for.

Let's see if this works.

So I think the answer to my question is positive.

Let's take an injective resolution $\mathcal{F}\rightarrow \mathcal{I}^\bullet$ of $\mathcal{F}$ and let $K_n^\bullet:=H^0_{Z_n}(X,\mathcal{I}^\bullet)$ be a complex which computes $H^*_{Z_n}(X,\mathcal{F})$. Then let $K_n^\bullet\rightarrow J_n^{\bullet,\bullet}$ be a Cartan–Eilenberg resolution of $K_n^\bullet$, such that $K^p\rightarrow J^{p,\bullet}$ is an injective resolution. Then, by definition, $$H^*(\operatorname{Tot}(\varprojlim_{n}J_n^{\bullet,\bullet}))=H^*(R\varprojlim_nK_n^\bullet).$$

The double complex $\varprojlim_n J_n^{\bullet,\bullet}$ induces by the two standard filtrations the mentioned hypercohomology spectral sequences

\begin{align} {}^IE_1^{pq}&={\varprojlim_n}^{(q)}H^0_{Z_n}(X,\mathcal{I}^p) \Rightarrow H^{p+q}(R\varprojlim_nK_n^\bullet), \\ {}^{II}E_2^{pq}&={\varprojlim_n}^{(p)}H^q\big(H^0_{Z_n}(X,\mathcal{I}^\bullet)\big) \Rightarrow H^{p+q}(R\varprojlim_nK_n^\bullet). \end{align}

By the arguments from the Proof of Proposition 4 in Section 2 of Schneider and Stuhler - The cohomology of $p$-adic symmetric spaces, we know

\begin{gather} {}^IE_1^{pq}=0 \text{ for } q \neq 0, \tag{1}\label{1} \\ {}^{II}E_2^{pq}=0 \text{ for } p \neq 0,1. \tag{2}\label{2} \end{gather}

Hence, both spectral sequences degenerate on the second page.

Additionally, let us fix $n \in \mathbb{N}$ and set

\begin{align} {}^IE_0^{\prime pq}&=J_n^{p,q} \Rightarrow H^{p+q}\bigl(\operatorname{Tot}(J_n^{\bullet,\bullet})\bigr)=H^{p+q}_{Z_n}(X,\mathcal{F}), \\ {}^{II}E_0^{\prime pq}&=J_n^{q,p} \Rightarrow H^{p+q}\bigl(\operatorname{Tot}(J_n^{\bullet,\bullet})\bigr)=H^{p+q}_{Z_n}(X,\mathcal{F}). \end{align}

Furthermore we have a natural morphism of double complexes

$$ \varprojlim_n J_n^{\bullet,\bullet} \rightarrow J_n^{\bullet,\bullet}.$$

As mentioned in EGA III, p. 30, a morphism of double complexes induces a natural morphism of spectral sequences (for both filtrations).

This implies we have a morphism $ {}^IE_1^{\bullet \bullet} \rightarrow {}^IE_1^{\prime\bullet \bullet}$ which is nothing but the morphism of complexes

$$ H^0_{Z}(X,\mathcal{I}^\bullet)={\varprojlim_n} H^0_{Z_n}(X,\mathcal{I}^\bullet) \hookrightarrow H^0_{Z_n}(X,\mathcal{I}^\bullet)$$ inducing the natural morphism $\varphi:H^*_Z(X,\mathcal{F})\rightarrow H^*_{Z_n}(X,\mathcal{F})$ in cohomology. The mentioned equality follows also by arguments from Schneider and Stuhler - The cohomology of $p$-adic symmetric spaces. Furthermore, the first spectral sequence implies $H^*_Z(X,\mathcal{F})=H^{*}(R\varprojlim_nK_n^\bullet).$

Then by the induced morphism on spectral sequences and the degeneration on the second page we have a commutative diagram

$\require{AMScd}$ \begin{CD} H^{n}(R\varprojlim_nK_n^\bullet) @>>> H^*_{Z_n}(X,\mathcal{F})\\ @V V V= @VV V\\ {}^{II}E_2^{0n} @>>> {}^{II}E_2^{\prime0n} \end{CD}

which is

$\require{AMScd}$ \begin{CD} H^*_Z(X,\mathcal{F})@>\varphi>> H^*_{Z_n}(X,\mathcal{F})\\ @V g V V= @VV \mathrm{id} V\\ {\varprojlim_n} H^*_{Z_n}(X,\mathcal{F})@>p_n>> H^*_{Z_n}(X,\mathcal{F}) \end{CD} what we were looking for.

Let's see if this works.

So I think the answer to my question is positive.

Lets take an injective resolution $\mathcal{F}\rightarrow \mathcal{I}^\bullet$ of $\mathcal{F}$ an let $K_n^\bullet:=H^0_{Z_n}(X,\mathcal{I}^\bullet)$ be a complex which computes $H^*_{Z_n}(X,\mathcal{F})$. Then be $K_n^\bullet\rightarrow J_n^{\bullet,\bullet}$ a Cartan-Eilenberg resolution of $K_n^\bullet$, such that $K^p\rightarrow J^{p,\bullet}$ is an injective resolution. Then, by definition, $$H^*(\mathrm{Tot}(\varprojlim_{n}J_n^{\bullet,\bullet}))=H^*(R\varprojlim_nK_n^\bullet).$$

The double complex $\varprojlim_n J_n^{\bullet,\bullet}$ induces by the two standard filtrations the mentioned hypercohomology spectral sequences

\begin{align} {}^IE_1^{pq}&={\varprojlim_n}^{(q)}H^0_{Z_n}(X,\mathcal{I}^p) \Rightarrow H^{p+q}(R\varprojlim_nK_n^\bullet), \\ {}^{II}E_2^{pq}&={\varprojlim_n}^{(p)}H^q\big(H^0_{Z_n}(X,\mathcal{I}^\bullet)\big) \Rightarrow H^{p+q}(R\varprojlim_nK_n^\bullet). \end{align}

By the arguments from the Proof of Proposition 4 in Section 2 of https://ivv5hpp.uni-muenster.de/u/pschnei/publ/pap/xsymm.pdf, we know

$${}^IE_1^{pq}=0 \text{ for } q \neq 0, \,\,\,\,\,\,\,\,\,(1)$$ $$ {}^{II}E_2^{pq}=0 \text{ for } p \neq 0,1. \,\,\,(2)$$

Hence, both spectral sequences degenerate on the second page.

Additionally, let us fix $n \in \mathbb{N}$ and set

\begin{align} {}^IE_0^{'pq}&=J_n^{p,q} \Rightarrow H^{p+q}\big(\mathrm{Tot}(J_n^{\bullet,\bullet})\big)=H^{p+q}_{Z_n}(X,\mathcal{F}), \\ {}^{II}E_0^{'pq}&=J_n^{q,p} \Rightarrow H^{p+q}\big(\mathrm{Tot}(J_n^{\bullet,\bullet})\big)=H^{p+q}_{Z_n}(X,\mathcal{F}). \end{align}

Furthermore we have a natural morphism of double complexes

$$ \varprojlim_n J_n^{\bullet,\bullet} \rightarrow J_n^{\bullet,\bullet}.$$

As mentioned in EGA III, p. 30, a morphism of double complexes induces a natural morphism of spectral sequences (for both filtrations)

This implies we have a morphism $ {}^IE_1^{\bullet \bullet} \rightarrow {}^IE_1^{'\bullet \bullet}$ which is nothing but the morphism of complexes

$$ H^0_{Z}(X,\mathcal{I}^\bullet)={\varprojlim_n} H^0_{Z_n}(X,\mathcal{I}^\bullet) \rightarrow H^0_{Z_n}(X,\mathcal{I}^\bullet)$$ inducing the natural morphism $\phi:H^*_Z(X,\mathcal{F})\rightarrow H^*_{Z_n}(X,\mathcal{F})$ in cohomology. The mentioned equality follows also by arguments from https://ivv5hpp.uni-muenster.de/u/pschnei/publ/pap/xsymm.pdf. Furthermore, the first spectral sequence implies $H^*_Z(X,\mathcal{F})=H^{*}(R\varprojlim_nK_n^\bullet).$

Then by the induced morphism on spectral sequences and the degeneration on the second page we have a commutative diagram

$\require{AMScd}$ \begin{CD} H^{n}(R\varprojlim_nK_n^\bullet) @>>> H^*_{Z_n}(X,\mathcal{F})\\ @V V V= @VV V\\ {}^{II}E_2^{0n} @>>> {}^{II}E_2^{'0n} \end{CD}

which is

$\require{AMScd}$ \begin{CD} H^*_Z(X,\mathcal{F})@>\varphi>> H^*_{Z_n}(X,\mathcal{F})\\ @V g V V= @VV id V\\ {\varprojlim_n} H^*_{Z_n}(X,\mathcal{F})@>p_n>> H^*_{Z_n}(X,\mathcal{F}) \end{CD} what we were looking for.

Let's see if this works.

So I think the answer to my question is positive.

Let's take an injective resolution $\mathcal{F}\rightarrow \mathcal{I}^\bullet$ of $\mathcal{F}$ an let $K_n^\bullet:=H^0_{Z_n}(X,\mathcal{I}^\bullet)$ be a complex which computes $H^*_{Z_n}(X,\mathcal{F})$. Then be $K_n^\bullet\rightarrow J_n^{\bullet,\bullet}$ a Cartan-Eilenberg resolution of $K_n^\bullet$, such that $K^p\rightarrow J^{p,\bullet}$ is an injective resolution. Then, by definition, $$H^*(\mathrm{Tot}(\varprojlim_{n}J_n^{\bullet,\bullet}))=H^*(R\varprojlim_nK_n^\bullet).$$

The double complex $\varprojlim_n J_n^{\bullet,\bullet}$ induces by the two standard filtrations the mentioned hypercohomology spectral sequences

\begin{align} {}^IE_1^{pq}&={\varprojlim_n}^{(q)}H^0_{Z_n}(X,\mathcal{I}^p) \Rightarrow H^{p+q}(R\varprojlim_nK_n^\bullet), \\ {}^{II}E_2^{pq}&={\varprojlim_n}^{(p)}H^q\big(H^0_{Z_n}(X,\mathcal{I}^\bullet)\big) \Rightarrow H^{p+q}(R\varprojlim_nK_n^\bullet). \end{align}

By the arguments from the Proof of Proposition 4 in Section 2 of https://ivv5hpp.uni-muenster.de/u/pschnei/publ/pap/xsymm.pdf, we know

$${}^IE_1^{pq}=0 \text{ for } q \neq 0, \,\,\,\,\,\,\,\,\,(1)$$ $$ {}^{II}E_2^{pq}=0 \text{ for } p \neq 0,1. \,\,\,(2)$$

Hence, both spectral sequences degenerate on the second page.

Additionally, let us fix $n \in \mathbb{N}$ and set

\begin{align} {}^IE_0^{'pq}&=J_n^{p,q} \Rightarrow H^{p+q}\big(\mathrm{Tot}(J_n^{\bullet,\bullet})\big)=H^{p+q}_{Z_n}(X,\mathcal{F}), \\ {}^{II}E_0^{'pq}&=J_n^{q,p} \Rightarrow H^{p+q}\big(\mathrm{Tot}(J_n^{\bullet,\bullet})\big)=H^{p+q}_{Z_n}(X,\mathcal{F}). \end{align}

Furthermore we have a natural morphism of double complexes

$$ \varprojlim_n J_n^{\bullet,\bullet} \rightarrow J_n^{\bullet,\bullet}.$$

As mentioned in EGA III, p. 30, a morphism of double complexes induces a natural morphism of spectral sequences (for both filtrations)

This implies we have a morphism $ {}^IE_1^{\bullet \bullet} \rightarrow {}^IE_1^{'\bullet \bullet}$ which is nothing but the morphism of complexes

$$ H^0_{Z}(X,\mathcal{I}^\bullet)={\varprojlim_n} H^0_{Z_n}(X,\mathcal{I}^\bullet) \hookrightarrow H^0_{Z_n}(X,\mathcal{I}^\bullet)$$ inducing the natural morphism $\varphi:H^*_Z(X,\mathcal{F})\rightarrow H^*_{Z_n}(X,\mathcal{F})$ in cohomology. The mentioned equality follows also by arguments from https://ivv5hpp.uni-muenster.de/u/pschnei/publ/pap/xsymm.pdf. Furthermore, the first spectral sequence implies $H^*_Z(X,\mathcal{F})=H^{*}(R\varprojlim_nK_n^\bullet).$

Then by the induced morphism on spectral sequences and the degeneration on the second page we have a commutative diagram

$\require{AMScd}$ \begin{CD} H^{n}(R\varprojlim_nK_n^\bullet) @>>> H^*_{Z_n}(X,\mathcal{F})\\ @V V V= @VV V\\ {}^{II}E_2^{0n} @>>> {}^{II}E_2^{'0n} \end{CD}

which is

$\require{AMScd}$ \begin{CD} H^*_Z(X,\mathcal{F})@>\varphi>> H^*_{Z_n}(X,\mathcal{F})\\ @V g V V= @VV id V\\ {\varprojlim_n} H^*_{Z_n}(X,\mathcal{F})@>p_n>> H^*_{Z_n}(X,\mathcal{F}) \end{CD} what we were looking for.