Universal norming functional in a l2 (AKA a separable inf-dim Hilbert space) Consider the space $\ell_2$.  We will use the notation that for any $x=(a_n)_{n=1}^\infty\in\ell_2$ and any $N\in\mathbb{N}$, the vector $x|_{[N,\infty)}$ denotes the sequence $(a_{N+n})_{n=0}^\infty$, and the vector $x^*$ denotes the decreasing rearrangement of $(|a_n|)_{n=1}^\infty$.
Definition 1.  Let $\mathbb{W}$ denote the set of all $\ell_\infty$-normalized, decreasing, positive, real-valued sequences in $c_0\setminus\ell_1$.
I want to construct a vector $w\in\mathbb{W}$ with some special properties.
Definition 2.  If $w\in\mathbb{W}$, we say that a subset $A$ of $\mathbb{R}^\mathbb{N}$ is called $\boldsymbol{w}$-almost-lengthwise-bounded if and only if both of the following conditions hold:
(1)  there is $C\in[1,\infty)$ (depending only on $A$) such that $x^*\cdot w\leqslant C$ for all $x\in A$; and
(2)  for each $\varepsilon>0$ there is $N\in\mathbb{N}$ (depending only on $A$ and $\varepsilon$) such that $x^*|_{[N,\infty)}\cdot w\leqslant\varepsilon$ for all $x\in A$.
Conjecture 1.  There exists $w\in\mathbb{W}$ such that for each $A\subseteq\mathbb{R}^\mathbb{N}$, the following are equivalent:
(i) $A$ is $w$-almost-lengthwise bounded
(ii) there exists $K>0$ (depending only on $A$) such that $\|x\|_{\ell_2}\leq K$ for all $x\in A$.
My ideas.
None, lol.  I mean, it seems like we should be able to take a countable dense subset of $S_{\ell_2}^*$ (the positive decreasing vectors in the unit sphere) and maybe concatenate some scaled norming vectors of finite support.  Or something along those lines.  But I can't get anything to work.
Note that the conjecture may not even be true.
 A: (I'm sorry, I'm going to use $\tilde{x}$ to denote the decreasing rearrangment of $x$, because $x^*$ screams "linear functional norming $x$" to me.)
Define $T_p:l^1\to\mathbb{R}^{\mathbb{N}}$; $(T_pw)_j=\sqrt[p]{j}w_j$ and $X_p=T_p^{-1}\mathbb{W}$.  Clearly, $X_p\neq\emptyset$ (for $p<\infty$).  Likewise, let $v:(1,\infty)\to\mathbb{W}$; $v(p)_j=j^{\frac{1}{p}-1}$.
Let (ii$p$) denote "$\sup_{x\in A}{\|x\|_p}<\infty$", which reduces to your original condition when $p=2$.  For any $p\leq q$, (ii$p$) implies (ii$q$), but the reverse is false.  Likewise, let (i$w$) denote "$A$ is $w$-almost-lengthwise-bounded".  Your claim is then: there exists $w\in\mathbb{W}$ such that (i$w$) iff (ii2).
Claim. (i$v(p)$) implies (ii$p$).
Proof.  Essentially, this follows from the answer to your other question: if $A$ is $v(p)$-almost-lengthwise-bounded and $x\in A$, then $$\|x\|_p=\|\tilde{x}\|_p\leq\langle v(p),\tilde{x}\rangle<\infty$$ uniformly in $x$.  QED.
Claim. Fix any $w\in X_p$.  Then (ii$p$) implies (i$T_pw$).
Proof.  By rescaling $A$, we might as well assume $\sup_{x\in A}{\|x\|_p}=1$.  Since the claim only gets stronger as we increase $A$, it suffices to take $A=\overline{B_{l^p}}(0,1)$; that is, the whole closed unit ball.
Consider $\tilde{A}$; that is, the decreasing rearrangements of $A$.  For any $x\in\tilde{A}$ and $J\in\mathbb{N}$, we have $$1\geq\sum_{j=1}^{\infty}{x_j^p}\geq\sum_{j=1}^J{x_j^p}\geq Jx_J^p$$  Thus $\sqrt[p]{J}x_J\leq1$.  So $$\langle T_pw,x\rangle=\sum_{j=1}^{\infty}{w_j\sqrt[p]{j}x_j}\leq\sum_{j=1}^{\infty}{w_j}\leq1$$ where the sums converge uniformly in $x$.  QED.
