A more general result can be given as follows:
$$ \sum_{i=1}^n \phi \left (\frac{\ f(a_i)}{\sum_{i=1}^n f (a_i)} \right ) \le \sum_{i=1}^n \phi \left (\frac{a_i}{\sum_{i=1}^n a_i} \right ) \tag {1} $$ where $\phi$ is an arbitrary convex function, $a_1,\dots, a_n \ge 0$ are given numbers for which $\sum_{i=1}^n a_i >0$ and $\sum_{i=1}^n f(a_i) >0$, $f: \mathbb R_{\ge 0} \to \mathbb R_{\ge 0}$ is an increasing function with $\frac{f(x)}{x}$ being decreasing on an interval including $a_1,\dots, a_n$. When $\frac{f(x)}{x}$ is strictly decreasing and $\phi$ is strictly convex, the equality holds only for $a_1=\dots=a_n$.
Examples of $f$ that can be used above are $f(x)=x^\theta, x\in \mathbb R$ for any $\theta \in (0,1)$, any increasing non-negative concave function, and $f(x)=-x\log x , x\in [0,\exp(-1)]$.
In particular, for $\phi(x)=\left |x -\frac{1}{n}\right |^\alpha$ with $\alpha \ge 1$, the above implies
$$\left \|\frac{(f (p_1),\dots,f (p_n))}{\sum_{i=1}^n f (p_i)} -\frac{1}{n} e \right \|_\alpha \le \left \|p -\frac{1}{n} e \right \|_\alpha \tag {2} $$ for $\alpha\ge 1$, $p_1,\dots,p_n \ge 0$ with $\sum_{i=1}^n p_i=1$, and any increasing function $f:I\to \mathbb R_{\ge 0}$ for which $\frac{f(x)}{x}$ is decreasing, where $I$ is the interval $[p_{(n)},p_{(1)}]$.
Considering $f(x)=-x\log x$, the above proves the inequality given in the OP when all probabilities $p_1,\dots,p_n$ are less than $\exp(-1)$ since $-x\log x$ is increasing only on the interval $[0,\exp(-1)]$. Solving the general case remains as open problem. Hope someone can help to solve it generally, which may need to use something other than majorization.
It worth noting that we have the following related result (with a key difference on the required condition for $f$), which follows from an existing result (see F4):
$$ \sum_{i=1}^n \phi \left (\frac{\ f(a_i)}{\sum_{i=1}^n f (a_i)} \right ) \ge \sum_{i=1}^n \phi \left (\frac{a_i}{\sum_{i=1}^n a_i} \right ) \tag {3} $$ where $\phi$ is an arbitrary convex function, $a_1,\dots, a_n \ge 0$ are given numbers for which $\sum_{i=1}^n a_i >0$ and $\sum_{i=1}^n f(a_i) >0$, $f: \mathbb R_{\ge 0} \to \mathbb R_{\ge 0}$ is a function with $\frac{f(x)}{x}$ being increasing on an interval including $a_1,\dots, a_n$. When $\frac{f(x)}{x}$ is strictly increasing and $\phi$ is strictly convex, the equality holds only for $a_1=\dots=a_n$.
Proof of (1)
Following the idea of using majorization suggested by @Ѕᴀᴀᴅ, a proof can be developed based on the following two facts:
F1: $x$ majorizes $y$ iff $\sum_{i=1}^n \phi(x_i) \ge \sum_{i=1}^n \phi(y_i)$ for any convex function $\phi$ (a well-known equivalent condition for majorization).
F2: Given $a_1\ge \dots \ge a_n \ge 0$ and $b_1\ge \dots \ge b_n \ge 0$ with $\frac{b_n}{a_n} \ge \dots \ge \frac{b_1}{a_1}$ where $ \sum_{i=1}^n a_i>0$ and $ \sum_{i=1}^n b_i >0$, vector $\frac{a}{\sum_{i=1}^n a_i}$ majorizes vector $\frac{b}{\sum_{i=1}^n b_i}$.
In F2, we adopt the convention $\frac{c}{0}=\infty$ for any $c\ge 0$, for the sake of simplicity.
For any given vector $u\ge 0$, if $w$ is defined as $w_i=f(u_i), i=1,\dots,n$ where $f: \mathbb R_{\ge 0} \to \mathbb R_{\ge 0}$ is an increasing function for which $\frac{f(x)}{x}$ is decreasing, then vectors $a=(u_{(1)},\dots, u_{(n)})$ and $b=(w_{(1)},\dots, w_{(n)})$ satisfy the condition of F2 as the sequence $\frac{f(u_{(i)})}{u_{(i)}}, i=1,\dots,n$ is increasing, and thus we obtain the following interesting result:
F3: If vector $u \ge 0$ and $f: \mathbb R_{\ge 0} \to \mathbb R_{\ge 0}$ is increasing with $\frac{f(x)}{x}$ being decreasing where $ \sum_{i=1}^n u_i>0$ and $ \sum_{i=1}^n f(u_i)>0$, vector $\frac{u}{\sum_{i=1}^n u_i}$ majorizes vector $\frac{(f (u_1),\dots,f (u_n))}{\sum_{i=1}^n f(u_i)}$.
A related result first appeared in page 4 of this 1967 paper [1, 2] (see also B.2. Proposition in page 188 of this book [3]) as follows:
F4: Given vector $u \ge 0$ and function $f: \mathbb R_{\ge 0} \to \mathbb R_{\ge 0}$ with $\frac{f(x)}{x}$ being increasing where $ \sum_{i=1}^n u_i>0$ and $ \sum_{i=1}^n f(u_i)>0$, vector $\frac{(f (u_1),\dots,f (u_n))}{\sum_{i=1}^n f(u_i)}$ majorizes vector $\frac{u}{\sum_{i=1}^n u_i}.$
To prove the above, again we can use F2 by setting $a=(w_{(1)},\dots, w_{(n)})$ and $b=(u_{(1)},\dots, u_{(n)})$ where $w_i=f(u_i), i=1,\dots,n$ for $u\ge 0$. In this case, we do not need to assume that is increasing because $b_1\ge \dots \ge b_n$ and $\small \frac{a_1}{b_1}\ge \dots \ge \frac{a_n}{b_n}$ together imply $a_1\ge \dots \ge a_n$.
Now one can see that the inequality (1) follows from F1 and F3, and (3) follows from F1 and F4.
Proof of F2
To be self-contained, here I provide the proof of F2, adjusted from the one given in page 3 of [1] for $a,b>0$. It is enough to show that for any $k=1,\dots,n$:
$$\small \frac{\sum_{i=1}^k a_i}{\sum_{i=1}^n a_i} \ge \frac{\sum_{i=1}^k b_i}{\sum_{i=1}^n b_i},$$
which is equivalent to
$$\small \left (\sum_{i=1}^k a_i \right) \left (\sum_{i=1}^n b_i \right)-\left (\sum_{i=1}^k b_i \right) \left (\sum_{i=1}^n a_i \right) \ge 0.$$
The above holds because the expression in the left side can be written as follows:
$$\small \left (\sum_{i=1}^k a_i \right) \left (\sum_{i=k+1}^n b_i \right)-\left (\sum_{i=1}^k b_i \right) \left (\sum_{i=k+1}^n a_i \right)=\sum_{i=1}^k \sum_{j=k+1}^n a_ia_j \left (\frac{b_j}{a_j}-\frac{b_i}{a_i} \right) \ge 0 $$
where we consider the simplifying conventions $0\times \infty=0$ and $\infty - \infty=0$.