What is a generalized limit?$\newcommand{\LIM}{\operatorname{LIM}}$
The question is about generalized limits $\LIM\colon\ell_\infty\to\mathbb R$. To deal with the question, we need to clearly define what we mean by a generalized limit.
We want $\LIM$ to be a linear continuous functional on $\ell_\infty$ which extends limit. I.e., if $x$ is a convergent sequence, then $\LIM x=\lim x$. Still, we need some additional condition - otherwise the claim in the question is not necessarily true.1 Let us check the linked blog (Wayback Machine). I will use the notation $\LIM$ as in the question, not $\widetilde\lim$ as in the blog
Using the Hahn-Banach theorem, one can extend the classical limit functional to generalised limit functionals $\LIM_{n \rightarrow \infty} a_n$, defined to be bounded linear functionals from the space $\ell^\infty({\bf N})$ of bounded real sequences to the real numbers ${\bf R}$ that extend the classical limit functional (defined on the space $c_0({\bf N}) + {\bf R}$ of convergent sequences) without any increase in the operator norm. ... It is not difficult to see that such generalised limit functionals will range between the limit inferior and limit superior. In fact, for any specific sequence $(a_n)_{n=1}^\infty$ and any number $L$ lying in the closed interval $[\liminf_{n \rightarrow \infty} a_n, \limsup_{n \rightarrow \infty} a_n]$, there exists at least one generalised limit functional $\LIM_{n \rightarrow \infty}$ that takes the value $L$ when applied to $a_n$.
Notice the phrase without any increase in the operator norm. So we only want to look at functionals such that $\|\LIM\|=1$. (It is easy to check that $\lim\colon c\to\mathbb R$ has norm equal to one.)
So let's take this as the definition of a generalized limit: $\LIM$ is a linear continuous functional defined on the space $\ell_\infty$ such that $\LIM$ extends the usual limit and $\|\LIM\|=1$.
Let us denote $$p(x)=\limsup x_n$$ for any $x\in\ell_\infty$. This will be the sublinear function that we'll later use in Hahn-Banach theorem. But first let us concentrate on the claim about limit inferior and limit superior. To this end, let us first show that if $\LIM$ is a generalized limit
$$(\forall x\in\ell_\infty) \LIM x\le p(x).$$
Let us show a weaker claim first, namely that $$\LIM x\le \limsup |x_n|\tag{*}$$
From the definition of generalized limit we have that $|\LIM x|\le \|x\|=\sup|x_n|$ and, consequently, also $\LIM x\le \sup|x_n|$.
Let us consider any bounded sequence $x$. Fix $k\in\mathbb N$. Let $y$ be the sequence which we obtain by changing the first $k$ terms of $x$ to zeroes. Then $\LIM x=\LIM y$. (From the fact that $x-y$ converges to zero, i.e., $\LIM(x-y)=\lim(x-y)=0$ and from linearity).
So we get $$\LIM x \le \sup |y_n| = \sup_{n>k} |x_n|.$$
Since this is true for every $k$ we get $$\LIM x \le \lim_{k\to\infty} \sup_{n>k} |x_n| = \limsup_{n\to\infty} |x_n|.$$
Now we want to use $(*)$ to get that also $$\LIM x\le\limsup x_n.$$
Let us consider any bounded sequence $x$. From boundedness we get that for some constant $K$ we have that $y=x+\overline K$ is positive (where $\overline K$ denotes the constant sequence). For a positive sequence we have $\LIM y \le\limsup |y_n|=\limsup y_n$, and thus we get
$$K+\LIM x = \LIM(x+\overline K) \le \limsup(x_n+K) = K+\limsup x_n$$
which implies that $\LIM x\le\limsup x_n$.
So we got that $\LIM x\le p(x)=\limsup x$. Using linearity and the fact that $-p(-x)=\liminf x$ we get from $-p(-x) \le \LIM x \le p(x)$ that
$$\liminf x \le \LIM x \le \limsup x.$$
So we gave some justification for this claim mentioned in the blog post: "such generalised limit functionals will range between the limit inferior and limit superior."
Of course, we could take this as a part of the definition. (Or take into the definition that condition that $\LIM$ is positive; i.e. $x\ge0$ implies $\LIM x\ge 0$.) But it seems natural to follow the definition mentioned in the linked blog post.
Every value is attained
So we now know that a generalized limit can only attain values between $\liminf x$ and $\limsup x$. How to show that for a given $x$ and any $L$ in this interval there is a generalized limit $\LIM$ such that $\LIM x=L$?
It is not difficult to see that if we apply Hahn-Banach Theorem to the limit function and the sublinear function $p(x)=\limsup x$, then this gives us a generalized limit. Moreover, if $p$ fulfils some additional conditions, we get that every value between $-p(-x)$ and $p(x)$ is attained for some extension.
Such version of Hahn-Banach theorem is formulated here: Reference for the range of possible values in Hahn-Banach Theorem.
Anyway, let us repeat the relevant part of the proof of Hahn-Banach theorem at least for the special case when $p(x)$ has all nice properties which we need here. (So we give a proof without relying on the theorem in the linked post.)
I will just remind that we have that $\limsup(x_n+y_n)\le \limsup x_n+\limsup y_n$ and $\limsup(cx_n)=c\limsup x_n$ for any $c>0$. (I.e., $p(x)$ is sublinear and positively homogeneous.) Moreover, for any convergent sequence $(y_n)$ we have $\limsup(x_n+y_n)=\limsup x_n + \lim y_n$.
We will see that these are precisely the properties needed here.
Extension to the dimension one higher
Lemma. Let $X$ be a vector space and $f$ be a linear function defined on a subspace $M\subseteq X$. Let $p\colon X\to\mathbb R$ be a function such that
\begin{gather*}
(\forall x,y\in X) p(x+y)\le p(x)+p(y),\\
(\forall x\in X)(\forall c>0) p(cx)=cp(x),\\
(\forall x\in X)(\forall y\in M) p(x+y)=p(x)+f(y).
\end{gather*}
Let us assume $f(x)\le p(x)$ for each $x\in M$ (i.e., $f$ is dominated by $p$ on $M$). Let $x_0\in X$ and $-p(-x)\le\theta\le p(x)$. Then there exists a linear function $\widehat f$ defined on $\widehat M=\operatorname{span}(M\cup\{x_0\})$ such that
$$(\forall x\in \widehat M) f(x) \le p(x)$$
and $$f(x)=\theta.$$
Notice that the first two conditions simply say that $p$ is sublinear and positive homogenous. Of course, we can prove Hahn-Banach theorem also under weaker conditions on $p$; however, this leads to to a slightly more complicated description of the range of possible values - this can be seen in the linked post. Let us stick to the simpler case which is sufficient for our purposes. (It is also worth mentioning that this lemma is basically just isolating one step which is actually used in the standard proof of Hahn-Banach Theorem.)
Proof. W.l.o.g. let us assume $x_0\notin M$. Let us define $\widehat f$ by putting
$$\widehat f(x+cx_0) = f(x)+c\theta.$$
for $x\in M$ and $c\in\mathbb R$. This give a linear map defined on $\widehat M$. Clearly, we have $f(x_0)=\theta$. We need to show that it is dominated by $p$, i.e., that we have
$$\widehat f(x+cx_0) \le p(x+cx_0)$$
for every $x\in M$ and $c\in\mathbb R$.
We have $p(x+cx_0)=f(x)+p(cx_0)$. Now if $c\ge 0$, then we get $p(cx_0)=cp(x_0)\ge c\theta$. For $c\le0$ we get $p(cx_0)=(-c)p(-x_0)\ge(-c)(-\theta)=c\theta$. So in either case we have
$$c\theta\le p(cx_0)$$
and thus
\begin{align*}
f(x+cx_0)&=f(x)+c\theta\\
&\le f(x)+p(cx_0)\\
&=p(x+cx_0).
\end{align*}
This concludes the proof.$\square$
Conclusion.
In our situation, we have the subspace $M$ consisting of all convergent sequences and a linear functional $f(x)=\lim x$.
Let we fix some $x$ and choose $\theta$ between $-p(-x)=\liminf x$ and $p(x)=\limsup x.$
The above lemma gives us an extension $\widehat f$ defined on $\widehat M=\operatorname{span}(M\cup\{x\})$ which is still dominated by $p$ and which, moreover, fulfills $$\widehat f(x)=\theta.$$
Now if we apply Hahn-Banach theorem once again to $\widehat f$, we get a functional $\LIM\colon\ell_\infty\to\mathbb R$ which is a generalized limit and we also have $$\LIM(x)=\widehat f(x)=\theta,$$ since $\LIM$ is an extension of $\widehat f$.
If all generalized limits have the same value
If we already know the above results, the conclusion which you mentioned in the part (2) is easy. If $x$ is a sequence such that for any generalized limit we get the same value $L$, that means that $\liminf x=\limsup x=L$. And this is equivalent to saying that $x$ converges to $L$.
1One example can be found in one of the answers to this question: Continuous extension of the limit functional. It also deals with extensions of limit, but the question is different. EDIT: To make this a bit clearer, I made a separate post about this: Do we have $f(x)\le\limsup x_n$ for every functional extending limit?
Best Answer
One way to get $LIM(x)\le p(x)$ is as follows:
$LIM$ is a linear functional with norm $1,$ such that if $x=(x_n)_n$ is convergent then $LIM(x)=\lim_n x_n.$
Let $x'=(x'_n)_n$ where $x'_n=\|x\|+x_n,$ so $x'_n\ge 0$ for all $n.$
For any $r>0$ define $y(r)=(y_n(r))_n$ and $x''(r)=(x''_n(r))_n$ by letting $y_n(r)=x'_n$ and $x''_n(r)=0$ if $x'_n\ge \|x\|+p(x)+r;$ while letting $y_n(r)=0$ and $x''_n(r)=x'_n$ if $x'_n<\|x\|+p(x)+r.$
Now $x'=y(r)+x''(r),$ but $\{n:y_n(r)\ne 0\}$ is finite, so $LIM(y(r))=0.$ So $$\|x\|+LIM(x)=LIM(x')=LIM(y(r)+x''(r))=LIM(x''(r))\le$$ $$\le \|x''(r)\|\le \|x\|+p(x)+r.$$ The last inequality comes from $\forall n\;(\,0\le x''_n<\|x\|+p(x)+r\,).$
Hence $LIM(x)\le p(x)+r$ for all $r>0,$ so $LIM(x)\le p(x).$