I am certainly not the best person to answer this question, as I do not have much insight to share regarding how to approach this kind of problems. My only (fairly obvious) suggestion is to *estimate* the relevant quantities in any way possible. In this process, it can be very helpful to reduce the calculations to lower dimensions, and the Fubini theorem will be our friend here.$\newcommand{\norm}[1]{\lVert #1 \rVert}$$\newcommand{\abs}[1]{\lvert #1 \rvert}$$\newcommand{\suchthat}{\ : \ }$$\newcommand{\set}[1]{\left\lbrace #1 \right\rbrace}$$\newcommand{\NN}{\mathbb{N}}$$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\ball}[2][0]{{B_{#2}(#1)}}$$\newcommand{\unitball}{\ball{1}}$$\newcommand{\dd}{\:\mathrm{d}}$$\newcommand{\label}[1]{\rlap{\qquad\qquad \text{#1}}}$

For this problem, my single intuition is that when one of the points $p_i$ drifts away from the centre of the unit ball, then every other point $p_j$ for $j\neq i$ cannot access a certain portion of the unit ball which sits close to $-p_i \: /\norm{p_i}$. By adding sufficiently many points, this should make the ratio $\chi(n)$ go to zero as $n$ goes to infinity. I will try to make this intuition precise in the remainder of this answer.

**Notation and preliminaries**

Fix the dimension $d$ of the ambient Euclidean space. Let $\norm{\cdot}$ denote the Euclidean norm on $\RR^d$, and $\ball[x]{r} = \set{ p\in\RR^d \suchthat \norm{p - x} \lt r }$ the open ball in $\RR^d$ with radius $r$ and centre $x$. Further, let $\mu$ be the Lebesgue measure on $\RR^d$, and $\mu_n = \mu^{\times n}$ the product measure on $\bigl(\RR^d \bigr)^{\times n}$ (i.e. Lebesgue measure on $\RR^{d n}$).

As in the statement of the question, $K_n$ will denote the following subset of $\unitball^{\times n}$:
$$ K_n = \set{ (p_1, \ldots, p_n) \in \unitball^{\times n} \suchthat \forall_{i,j} \ \norm{p_i - p_j} \lt 1 } $$
We shall prove that
$$ \chi(n) = \frac{\mu_n(K_n)}{\mu_n \bigl( \unitball^{\times n} \bigr)} = \frac{\mu_n(K_n)}{\mu\bigl(\unitball\bigr)^n} $$
converges to $0$ as $n\to\infty$. In fact, I will give (fairly coarse) upper and lower bounds for $\chi(n)$. Nevertheless, I am sure one can give much stricter bounds for $\chi(n)$ and perhaps even describe its asymptotic behaviour.

**Upper bound for $\chi(n)$**

Fix a positive integer $n$. Define
$$ X_n = \set{ (p_1, \ldots, p_n)\in K_n \suchthat \forall_i \ \norm{p_i} \lt 1/2 } $$
(the choice of the number $1/2$ here is mostly arbitrary) and $Y_n = K_n\setminus X_n$ to be the complement of $X_n$ in $K_n$. The set $X_n$ is contained in $\ball{1/2}^{\times n}$ and thus
$$ \mu_n(X_n) \leq \mu_n\bigl(\ball{1/2}^{\times n}\bigr) = \mu\bigl(\ball{1/2}\bigr)^n $$
Now define for each $i\in\set{1, \ldots, n}$ the set
$$ Z_{n,i} = \set{ (p_1, \ldots, p_n)\in K_n \suchthat \norm{p_i} \geq 1/2 } $$
It is easy to check that
$$ Y_n = \bigcup_{i=1}^n Z_{n,i} $$
Moreover, by permuting the first and $i$-th points (which gives a measure preserving self-bijection of $K_n$), we see that $\mu_n(Z_{n,i}) = \mu_n(Z_{n,1})$ for all $i\in\set{1, \ldots, n}$. It follows that
$$ \mu_n(Y_n) \leq n \cdot \mu_n(Z_{n,1}) $$
and therefore
$$ \mu_n(K_n) = \mu_n(X_n) + \mu_n(Y_n) \leq \mu\bigl(\ball{1/2}\bigr)^n + n\cdot \mu_n(Z_{n,1}) \label{(A)} $$

Now we perform a crude estimation of the volume of $Z_{n,1}$. The Fubini theorem entails
$$ \mu_n(Z_{n,1}) = \int_{\unitball\setminus\ball{1/2}} \mu_{n-1}(P_x) \dd \mu(x) $$
where $P_x \subset K_{n-1}$ is the following cross-section of $Z_{n,1}$, for each $x\in\unitball\setminus\ball{1/2}$:
$$ P_x = \set{ (p_1, \ldots, p_{n-1}) \in \bigl(\RR^d \bigr)^{\times (n-1)} \suchthat (x, p_1, \ldots, p_{n-1}) \in K_n } $$

**Lemma:** For each $x\in\unitball\setminus\ball{1/2}$, the inequality $\mu_{n-1}(P_x) \leq \mu\bigl(\unitball\cap\ball[1/2,0,\ldots,0]{1}\bigr)^{n-1}$ holds.

I will present a proof of the lemma at the end of this answer. We now apply the estimate in the lemma to the preceding integral expression for $\mu_n(Z_{n,1})$ to obtain:
$$ \mu(Z_{n,1}) \leq \mu\bigl( \unitball \bigr) \cdot \mu\bigl( \unitball\cap\ball[1/2,0,\ldots,0]{1} \bigr)^{n-1} $$
Putting this together with estimate (A):
$$ \mu_n(K_n) \leq \mu\bigl(\ball{1/2}\bigr)^n + n\cdot \mu\bigl( \unitball \bigr) \cdot \mu\bigl(\unitball\cap\ball[1/2,0,\ldots,0]{1}\bigr)^{n-1} $$
and further using the fact that $\ball{1/2} \subset \unitball\cap\ball[1/2,0,\ldots,0]{1}$, we simplify
$$ \mu_n(K_n) \leq (n+1) \cdot \mu\bigl( \unitball \bigr) \cdot \mu\bigl(\unitball\cap\ball[1/2,0,\ldots,0]{1}\bigr)^{n-1} $$
Consequently, we obtain an upper bound for $\chi(n)$:

$$ \chi(n) \leq (n+1) \left( \frac{\mu\bigl(\unitball\cap\ball[1/2,0,\ldots,0]{1}\bigr)}{\mu\bigl(\unitball\bigr)} \right)^{n-1} = (n+1) \rho^{n-1} $$

where $0 \lt \rho \lt 1$. Note that $\rho$ depends only on $d$. In particular, $\chi(n)$ converges to zero as $n\to\infty$.

**Lower bound for $\chi(n)$**

It is very easy to give a crude lower bound for $\chi(n)$. Simply observe that $X_n \subset K_n$, and that $X_n = \ball{1/2}^{\times n}$ (here we *do* require the choice of $1/2$ in the definition of $X_n$). Therefore,
$$ \mu_n(K_n) \geq \mu_n(X_n) = \mu\bigl(\ball{1/2}\bigr)^n $$
and so

$$ \chi(n) \geq \left( \frac{\mu\bigl(\ball{1/2}\bigr)}{\mu\bigl(\ball{1}\bigr)} \right)^n = 2^{-dn} $$

**Proof of the lemma**

We make use of the rotational symmetry of $Z_{n,1}$. Choose a rotation on $\RR^d$ which takes the point $x$ to the point $(\norm{x},0,\ldots,0)$ on the first axis. Applying that rotation componentwise gives a measure preserving bijection between $P_x$ and $P_{(\norm{x},0,\ldots,0)}$, and we see that
$$ \mu_{n-1}(P_x) = \mu_{n-1}\bigl( P_{(\norm{x},0,\ldots,0)} \bigr) \label{(1)} $$

On the other hand, it is straightforward to check that
$$ P_x = \set{ (p_1, \ldots, p_{n-1}) \in K_{n-1} \suchthat \forall_i \ \norm{p_i - x} \lt 1 } $$
and it follows that
$$ P_x \subset \bigl(\unitball\cap\ball[x]{1}\bigr)^{\times (n-1)} \label{(2)} $$

*Claim*: For $0\leq s\leq t$, the following inclusion holds:
$$ \unitball\cap\ball[t,0,\ldots,0]{1} \subset \unitball\cap\ball[s,0,\ldots,0]{1} \label{(3)} $$

*Proof of claim:*

- For $a\in\RR$, we have $\abs{a-s} \leq \max\set{ \abs{a-t}, \abs{a} }$: either $a \lt s$ which implies $\abs{a-s} \leq \abs{a-t}$, or $a\geq s$ which implies $\abs{a-s}\leq \abs{a}$.
- Thus, for any $y\in\RR^d$ we have $\abs{y_1-s} \leq \max\set{ \abs{y_1-t}, \abs{y_1} }$ which by a simple calculation implies $\norm{y-(s,0,\ldots,0)} \leq \max\set{ \norm{y-(t,0,\ldots,0)}, \norm{y} }$.

*End of proof.*

The inclusions (2) and (3) prove
$$ P_{(t,0,\ldots,0)} \subset \bigl(\unitball\cap\ball[t,0,\ldots,0]{1}\bigr)^{\times (n-1)} \subset \bigl(\unitball\cap\ball[1/2,0,\ldots,0]{1}\bigr)^{\times (n-1)} $$
for $1/2 \leq t \lt 1$. Using (1), we conclude that for each $x\in\unitball\setminus\ball{1/2}$
$$ \mu_{n-1}(P_x) = \mu_{n-1}\bigl( P_{(\norm{x},0,\ldots,0)} \bigr) \leq \mu_{n-1}\Bigl(\bigl(\unitball\cap\ball[1/2,0,\ldots,0]{1}\bigr)^{\times (n-1)}\Bigr) $$
and so $\mu_{n-1}(P_x) \leq \mu\bigl(\unitball\cap\ball[1/2,0,\ldots,0]{1}\bigr)^{n-1}$ as desired.