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 $\int_1^{1+\delta} n^{1/p} dp >\delta n^{1+\delta}$ 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 $e_i$ 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 $l_1$ ?
Question 2: Is the space $l_1$ saturated?