= Solution
For $K\geq1$, define the normalized $\ell^1$ tree
$$
T_K(X)=\left\{(x_1,\ldots,x_n)\in S_X^{<\omega}:\frac1K\sum|a_i|\leq\left\|\sum a_ix_i\right\|\leq\sum|a_i|\text{ for all scalars }a_i\right\}.
$$
The upper bound follows from normalization. This is a closed tree: it suffices to impose the lower inequalities for a countable dense set of scalar tuples. The <Bourgain l1-index> is
$$
\boxed{I_1(X)=\sup_{K\in\mathbb N}o(T_K(X)),}
$$
where $o$ is its well-founded tree order, and the index is $\omega_1$ if a tree is ill-founded. An infinite branch is exactly a sequence equivalent to the unit vector basis of $\ell^1$.
If $X$ is separable and contains no $\ell^1$, its unit sphere is Polish and every $T_K(X)$ is closed and well-founded. Part (a) makes each order countable, and their countable supremum is countable. In particular every separable <reflexive Banach space> has countable index, since its closed subspaces are reflexive whereas $\ell^1$ is not.
We now construct separable reflexive spaces with arbitrarily high index. Take any countable well-founded rooted tree $T$. Recursively, at each node $t$, define
$$
X_t=\mathbb R\oplus_1\left(\bigoplus_{s\text{ an immediate child of }t}X_s\right)_{\ell^2}.
$$
At a terminal node this is just $\mathbb R$. The recursion is legitimate by well-foundedness. Each $X_t$ is separable and reflexive: a countable $\ell^2$ sum of separable reflexive spaces is separable and reflexive, and adjoining one dimension in a finite direct sum preserves reflexivity. Let $X_T$ be the root space, and let $e_t$ be the unit vector at each node coordinate.
Along every finite branch $t_1\prec\cdots\prec t_n$, the recursive norm is exactly
$$
\left\|\sum_{i=1}^na_ie_{t_i}\right\|=\sum_{i=1}^n|a_i|.
$$
Indeed, at every step only one child component is occupied, so its $\ell^2$ norm equals that component's norm, while the node coordinate adds by the $\ell^1$ sum. Thus $T$ embeds by extension into $T_1(X_T)$, and its rank is a lower bound for the index, up to the harmless empty-root convention. Countable well-founded trees have arbitrarily high countable ranks: add a root for a successor stage and attach countably many trees of cofinal ranks at a limit stage.
Suppose now a separable reflexive space $U$ were universal. Put $\beta=I_1(U)<\omega_1$ and choose $T$ of height larger than $\beta$. An isomorphic embedding $J:X_T\to U$ has bounds $c\|x\|\leq\|Jx\|\leq C\|x\|$. The normalized images $Je_t/\|Je_t\|$ along each branch satisfy the lower $\ell^1$ estimate $c/C$, so the tree embeds into $T_K(U)$ for some integer $K\geq C/c$. It would have order greater than $\beta$, contradicting its definition. Therefore \b[no separable reflexive Banach space is universal for all separable reflexive Banach spaces]. The embedding constant need not be uniform across spaces; the supremum over all $K$ handles this.
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