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Proof of Fact 4: We set $q_k = 1 + \frac{1}{2^k} $. Observe that for all $k \in N$
$sup \{ |x_n |_{q_k} : n\in N \} \leq \frac {1}{2^k} $.

$ \frac {|z_{l}|_{q_k }} {2^{k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^{k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$
where $z_l = \sum _{n\in F_{l}} x_n$.

$ \frac {\int_{q_k}^2 |z_l|_p dp } {\int_1^{2} |z_l|_p dp} < \frac {\int_{q_k}^2 |z_k|_p dp } {\int_1^{q_{k+1}} |z_k|_p dp} < \frac {|z_{l}|_{q_k }} {2^{k+1}|z_{l}|_{q_{k+1}} } \leq (\frac {|z_{l}|_{q_k }} {2^{k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^ {k+1}C_{k+1} l^{1/q_{k+1}}}) + \frac { C_{k} l^{1/{q_k}}} { 2^{k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$.

Proof of Fact 4: We set $q_k = 1 + \frac{1}{2^k} $. Observe that for all $k \in N$
$sup \{ |x_n |_{q_k} : n\in N \} \leq 2^k $.

$ \frac {|z_{l}|_{q_k }} {2^{-k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^{-k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$
where $z_l = \sum _{n\in F_{l}} x_n$.

$ \frac {\int_{q_k}^2 |z_l|_p dp } {\int_1^{2} |z_l|_p dp} < \frac {\int_{q_k}^2 |z_k|_p dp } {\int_1^{q_{k+1}} |z_k|_p dp} < \frac {|z_{l}|_{q_k }} {2^{-k+1}|z_{l}|_{q_{k+1}} } \leq (\frac {|z_{l}|_{q_k }} {2^{-k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^ {-k+1}C_{k+1} l^{1/q_{k+1}}}) + \frac { C_{k} l^{1/{q_k}}} { 2^{-k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$.

Edit 1 The answer to this question is also affirmative. In particular the following holds:

Fact 3: Every closed infinite dimensional subspace $Z$ of the space has a further subspace $Y$ which is complemented in the space and isomorphic to $l_1$.

To prove this we first observe that it is enough to consider block subspaces namely subspaces generated by a normalized block sequence $(x_n )_n$. For such a sequence we will show the following:

Fact 4: For every normalized block sequence $( x_n)_n$ there exists a sequence $(F_k )_k $ with $F_k \subset N$, $\#F_k = n_k$ such setting $z_k = \sum_{n\in F_k}x_n$ the following holds.

For every $q > 1 $ $ lim_k \frac {\int_1^{q} |z_k|_p dp } {\int_1^{2} |z_k|_p dp} = 1 $.

With this result and repeating the proof of Fact 2 above for the sequence $(z_k)_k $ we conclude that it has a subsequence equivalent to $l_1$ basis. moreover the subspace generated by a further subsequence is complemented in the space. This follows from the answer to Question 1 above with a small modification at the last part.

Proof of Fact 4: We set $q_k = 1 + \frac{1}{2^k} $. Observe that for all $k \in N$
$sup \{ |x_n |_{q_k} : n\in N \} \leq \frac {1}{2^k} $.

We assume that $ lim_n |x_n |_{q_k} = C_k $ for all $k\in N $ (otherwise we pass to a subsequence).
Observe that $(C_k)_k$ is increasing.

Claim: For every $k\in N$ there exists $F_k \subset N$ such that setting $z_k = \sum_ {n\in F_k} x_n $ we have that

$ \frac {\int_1^{q_k} |z_k|_p dp } {\int_1^{2} |z_k|_p dp} > 1- \frac{1} {k} $

Proof of the Claim: Since $ |x_n|_{q_k} \rightarrow C_k$ and $|x_n|_{q_{k+1}} \rightarrow C_{k+1}$ for every $l\in N$ we may select $F_l \subset N$ such that $\#F_l =l$ and

$ \frac {|z_{l}|_{q_k }} {2^{k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^{k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$
where $z_l = \sum _{n\in F_{l}} x_n$.

As in Fact 1 we have that:

$ \frac {\int_{q_k}^2 |z_l|_p dp } {\int_1^{2} |z_l|_p dp} < \frac {\int_{q_k}^2 |z_k|_p dp } {\int_1^{q_{k+1}} |z_k|_p dp} < \frac {|z_{l}|_{q_k }} {2^{k+1}|z_{l}|_{q_{k+1}} } \leq (\frac {|z_{l}|_{q_k }} {2^{k+1}|z_{l}|_{q_{k+1}} } - \frac { C_{k} l^{1/{q_k}}} { 2^ {k+1}C_{k+1} l^{1/q_{k+1}}}) + \frac { C_{k} l^{1/{q_k}}} { 2^{k+1}C_{k+1} l^{1/q_{k+1}}} \rightarrow 0$.

We choose $l_k$ such that $\frac {\int_{q_k}^2 |z_{l_k}|_p dp } {\int_1^{2} |z_{l_k}|_p dp} <\frac {1} { k}$ and we set $ F_k = F_{l_k} $ and $ z_k = z_{l_k} $. Clearly $F_k , z_k $ satisfy the conclusion of the claim.

Conclusion: The sequence $(z_k )_k $ satisfies:

For every $q > 1 $ $ lim_k \frac {\int_1^{q} |z_k|_p dp } {\int_1^{2} |z_k|_p dp} = 1 $.Hence as in Fact 2 there is a subsequence $(z_{k_m})_m$ equivalent to $l_1$ basis. This subsequence has a further subsequence with the property that the space that generates is complemented in the whole space.

For $x$ as above and $ p\in (1,2] $ we set $f_p^x = \frac{1} {(\sum_{i=1}^n \lambda^p)^1/q } \sum_{i=1}^n \lambda_i^{1/q-1} e_i$ where $ 1/p + 1/q = 1 $.
The functional $f^x_p$ is the unique normalized element of $l_q$ that norms $x$ as an element of $l_p $.
Observe that for a given $x$ as above and $z\in l_1$ the function $f_p ^x (z) $ with variable $ p \in (1,2]$ is continuous hence for a Borel $A \subset (1,2]$ the integral
$f_A ^x(z)= \int_A f_p^x (z) dp$
is well defined for all $z \in l_1$ and $f_A ^x$ is linear.

We set $A_k= (\delta_{k+1}, \delta_k)$ and $x_k = x_{n_k}$ as they appeared in Step 1 above. Then the sequence $(f_{A_k}^{x_k} )_k$ satisfies the previous requirements hence it has a subsequence equivalent to $c_o$ basis.

For $x$ as above and $ p\in (1,2] $ we set $f_p^x = \frac{1} {(\sum_{i=1}^n \lambda_i^p)^1/q } \sum_{i=1}^n \lambda_i^{1/q-1} e_i$ where $ 1/p + 1/q = 1 $.
The functional $f^x_p$ is the unique normalized element of $l_q$ that norms $x$ as an element of $l_p $.
Observe that for a given $x$ as above and $z\in l_1$ the function $f_p ^x (z) $ with variable $ p \in (1,2]$ is continuous hence for a Borel $A \subset (1,2]$ the integral
$f_A ^x(z)= \int_A f_p^x (z) dp$
is well defined for all $z \in l_1$ and $f_A ^x$ is linear.

We set $A_k= (\delta_{k+1}, \delta_k)$ and $x_k = x_{n_k}$ as they appeared in Step 1 above. Then the sequence $(f_{A_k}^{x_k} )_k$ satisfies $ |f_{A_k}^{x_k}|_* \geq \frac{1} {4}$ and $(A_k)_k$ are disjoint. Hence it has a subsequence equivalent to $c_o$ basis.