Everywhere in this post, $p$ is supposed to be in the range $\left[1,\infty\right)$. Recall that a Banach space $X$ is called an $\mathcal{L}_{p,1+}$-space if for every finite-dimensional subspace $E\subseteq X$ and every $\varepsilon>0$ there exists a finite-dimensional subspace $E\subseteq F\subseteq X$ which is $(1+\varepsilon)$-isomorphic to $\ell_p(n)$, where $n$ is the dimension of $F$.
Lindenstrauss and Pelczynski prove in their paper Absolutely summing operators in $\mathcal{L}_p$ spaces and their applications that $\mathcal{L}_{p,1+}$ spaces are isometrically $L_p(\mu)$ spaces. They mention that this implies that a Banach space $X$ is isometric to an $L_p(\mu)$ if and only if there exists a sequence (actually a net in general, but let's stick just to separable spaces where sequences suffice) of increasing, in inclusion, finite-dimensional Banach subspaces $(E_n)_n$ such that the union $\bigcup_n E_n$ is dense in $X$ and each $E_n$ is isometric to $\ell_p(k(n))$, where $k(n)$ is the dimension of $E_n$.
That's actually not clear to me. I guess it should be easy that an existence of such a sequence $(E_n)_n$ implies that $X$ is a $\mathcal{L}_{p,1+}$ space?
Let me ask something stronger: Is it true that a (separable) Banach space $X$ is a $\mathcal{L}_{p,1+}$ space (or isometric to an $L_p(\mu)$ space) if and only if there exists a sequence of increasing, in inclusion, finite-dimensional Banach subspaces $(E_n)_n$ such that again the union $\bigcup_n E_n$ is dense in $X$, but now each $E_n$ is $(1+\varepsilon_n)$-isomorphic to $\ell_p(k(n))$, where $k(n)$ is as before, and $\varepsilon_n\to 0$.