We will show that the space contains isomorphically the space $l_1$ therefore the space is not reflexive. We start with the following that I posed as a question in a previous comment.
Fact 1: For every $0<\delta < 1 $ $lim_n \frac{\int_1^{1+\delta} n^{1/p} dp } {\int_1^2 n^{1/p} dp } = 1$.
Proof: Since the function $n^{1/p}$, $1\leq p \leq 2$ is decreasing we have that $\int_1^{1+\delta} n^{1/p} dp >\delta n^{1/1+\delta}$ and
$n^ {1/1+2\delta} > \int_{1+2\delta}^2 n^{1/p} dp$
Now for $n\in N$ we have that $ \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^2 n^{1/p} dp } < \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^{1+\delta} n^{1/p} dp } < \frac{n^{1/1+2\delta }} {\delta n^{1/1+ \delta }} = \frac{1} { \delta} \frac {1}{n^{\delta/ (1+\delta)(1+2\delta)}}$.
Hence for every $0<\delta <1/2$
$lim_n \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^2 n^{1/p} dp } =0$ which finishes the proof of Fact 1.
Fact 2: We start with the following classical result.
If $(f_n)_n$ is a normalized sequence in $L^1 [1,2]$ which is not uniformly integrable (i.e. there exists $\epsilon>0$ such that for every $\delta>0$
there exists a Borel set $A$ with $\lambda (A)<\delta$ and $\int_A |f_n| > \epsilon$ for infinite $n\in N$ ) then $(f_n)_n$ has a subsequnce equivalent to $l_1$ basis.
This result is due to Kadec and Pelczynski ( see J. Diestel: Sequences and Series in Banach Spaces (Graduate Texts in Mathematics, 92) p. 93).
Next in the space we denote $(e_i)_i\in N$ the basis of $l_1$ which is a symmetric basis for the space.For $n\in N $ we set $z_n = \sum_ {i=1} ^{n} e_i$ and $x_n = \frac{1} {\int_1^2 n^{1/p} dp } z_n$ which has norm 1.
Consider the function $f_n(p) = |x_n|_p $ $1\leq p \leq 2 $ and Fact 1 yields that the sequence $(f_n)$ it is not uniformly integrable.Therefore $(f_n)$ has a subsequence equivalent to $l_1$ basis which implies that $(x_n)$ satisfies the same property in the norm of the space. We will adapt Kadec - Pelczynski's proof in the setting of $(x_n)$.
Step 1 : There exists a decreasing sequence $(\delta_k )$ and a subsequence $(x_{n_k})$ such that for all $k$ we have that $\int_{1+\delta_{k+1}}^{1+\delta_k} |{x_{n_k}}|_p dp > \frac{1} { 4}$.
The proof uses induction and the following :
From Fact 1 for $ 0< \delta < 1$ there exists $n\in N$ such that $\int_1^{1+\delta}| {x_n}|_p dp > \frac{1} {2}$. For this $n$ there exists $\delta_1 < \delta $ such that $\int_1^{1+\delta_1}|{x_n}|_p dp < \frac{1} {4}$. Hence $\int_{1+\delta_1}^{1+\delta} |{x_n}|_p dp > \frac{1} {4}$.
Step 2: We set $A_k = [\delta_{k+1}, \delta_{k }]$. There exists an infinite $I \subset N $ such that for every $k\in I$ setting $ B_k = \cup \{ A_j : j\in I, j\neq k \} $ we have that $\int_{B_k} |{x_{n_k}|_p}dp < \frac {1} {8}$.
This is a classical result due to H. P. Rosenthal and an elegant and short proof was given by J. Kupka ( see Page 82 in the aforementioned reference). We assume that $ I = N $
Claim: The sequence $(x_{n_k} )_{k\in N}$ is equivalent to $l_1$ basis.
Indeed
$\int_1^2 |\sum_{j=1}^k \lambda_{j} x_{n_j} |_p dp > \int_{\cup_{i\in N} A_i} |\sum_{j=1}^k \lambda_{j} x_{n_j} |_pdp \geq \sum_{j=1} ^{k} (\int_{A_j} |\lambda_{j} x_{n_j}|_pdp - \int_{B_j} |\lambda_{j} x_{n_j}|_p dp)\geq
\frac {1}{8} \sum_{j=1}^{k} |\lambda_j|$.
I have two questions related to this result.
Question 1 : Does the space contain a complemented subspace isomorphic to $l_1$ ?
Edit: The answer to Question 1 is affirmative hence the dual of the space contains isomorphically the space $l_\infty $.
The functionals $f_A^x $ where A is a Borel subset of [1,2] and $x=\sum_{i=1}^n \lambda_i e_i (\lambda_i \geq 0). $
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.
Properties of $f_A ^x$.
We denote by $|.|$ the norm of the space and by $|.|_*$ the norm of its dual.
For all $x$ , $A$ $| f_A ^x |_* \leq 1$ moreover if $ |x|\leq 1$ and $\int_A |x|_p dp \geq c > 0 $ then $| f_A ^x |_* \geq c$.
For $(A_k)_{k=1}^m $ disjoint Borel sets , $(x_k)_{k=1}^m$ in $l_1$ and $(\alpha_k)_{k=1}^m$ reals we have that.
$ | \sum_{k=1}^{m} \alpha_{k}f_{A_k}^{x_k} |_* \leq max \{|\alpha_k| : k=1,...,m \}$
In particular every sequence $(f_{A_k}^{x_k} )_k$ with $(A_k))_k$ disjoint Borel sets is weakly null since every n-average of them has norm less or equal to $1/n$.
Furthermore if for every $k \in N$ $|f_{A_k}^{x_k}|_*\geq c >0 $ then $(f_{A_k}^{x_k} )_k$ has a subsequence which is Schauder basic which yields that this subsequence is equivalent to $c_o$ basis.
The dual of the space contains isomorphically $c_0$.
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.
The space has a complemented subspace isomorphic to $l_1$.
This is an immediate consequence of the previous result. A classical Theorem states that if $c_0$ is isomorphic to a subspace of $X^*$ then $l^1$ is isomorphic to a complemented subspace of $X$.
Question 2: Is the space $l_1$ saturated?