Let $\lambda$ be a Young tableaux and $V_{\lambda}$ the standard simple $\mathbb{C}[S_n]$-module associated to it, constructed as $\mathbb{C}[S_n]c_{\lambda}$.
We denote by $x=\sum\limits_{1\leq i <j\leq n}(i\;\; j)$. I should demonstrate that multiplicating on the left by $x$, corresponds on $V_{\lambda}$, to $ c(\lambda)\,\text{id}$, where $$c(\lambda)=\sum_{j} \sum_{i=1}^{\lambda_j}(i-j)\,.$$
Now, $x \in Z\big(\mathbb{C}[S_n]\big)$, so multiplicating on the left is an $S_n$-equviariant endomorphism of $V_{\lambda}$ and $V_{\lambda}$ being simple, it corresponds to multiplication by a scalar. But I really do not know how to find this scalar.
For a fixed partition $\lambda=\left(\lambda_1,\lambda_2,\ldots,\lambda_l\right)$ of $n$, let $t$ be a Young tableau of shape $\lambda$. Let $R_t\leq S_n$ and $C_t\leq S_n$ denote the row stabilizer and the column stabilizer of $t$, respectively. Write $\Omega_t$ for the orbit of $t$ under $R_t$. Then, recall that $C_t$ acts on $\Omega_t$ via the natural action inherited from $S_n$: $$h\cdot \Omega_t:=\big\{h\cdot s\,\big|\,s\in \Omega_t\big\}\text{ for each }h\in C_t\,.$$ Observe that $\sigma\cdot \Omega_t=\Omega_{\sigma\cdot t}$ for all $\sigma\in S_n$. Now, let $\Gamma_t$ be the orbit of $\Omega_t$ under $C_t$. In this problem, we may take $t$ to be the Young tableau whose $k$-th column is given by $$\left(\sum_{r=1}^{k-1}\,\lambda_r+1,\sum_{r=1}^{k-1}\,\lambda_r+2,\ldots,\sum_{r=1}^{k}\,\lambda_r-1,\sum_{r=1}^{k}\,\lambda_r\right)\,,$$ although a specific choice of $t$ is irrelevant to the proof.
We can view the Specht module $V_\lambda$ to be $\mathbb{C}[S_n]\cdot \left(\kappa_t\cdot \Omega_t\right)$, where $$\kappa_t:=\sum_{h\in C_t}\,\text{sign}(h)\,h\,.$$ Write $e_t:=\kappa_t\cdot\Omega_t$. Note that $e_t$ is a linear combination of elements of $\Gamma_t$ with coefficients $\pm1$. Recall that $x=\sum\limits_{1\leq i<j\leq n}\,(i\;\;j)=\sum\limits_{\tau\in T}\,\tau$, where $T$ is the set of all transpositions in $S_n$, is a central element of $\mathbb{C}[S_n]$. Therefore, $x$ acts on $V_\lambda$ by a scalar multiplication $c(\lambda)$. It suffices to determine $c(\lambda)$ from $x\cdot e_t$.
Now, since $x\cdot e_t$ is a scalar multiple of $e_t$, we may only look at terms in $x\cdot e_t$ that belong in $\Gamma_t$. Observe that, for $h\in C_t$ and $\tau \in T$, $\tau\cdot (h\cdot \Omega_t)=\tau\cdot \Omega_{h\cdot t}$ is in $\Gamma_t$ if and only if $\tau\in R_{h\cdot t}$ or $\tau\in C_t$. In other words, $$\begin{align}x\cdot e_t&=\sum_{\tau\in T}\,\tau\cdot\left(\sum_{h\in C_t}\,\text{sign}(h)\,(h\cdot \Omega_t)\right)=\sum_{h\in C_t}\,\sum_{\tau \in T}\,\text{sign}(h)\,\tau\cdot (h\cdot \Omega_t)\\&=\sum_{h\in C_t}\,\sum_{\tau \in T\cap R_{h\cdot t}}\,\text{sign}(h)\,\big(\tau\cdot(h\cdot \Omega_t)\big)+\sum_{h\in C_t}\,\sum_{\tau \in T\cap C_t}\,\text{sign}(h)\,\big(\tau\cdot(h\cdot \Omega_t)\big)\,.\end{align}$$
Because $h\cdot \Omega_t=\Omega_{h\cdot t}$ and $C_{h\cdot t}=C_t$ for every $h\in C_t$, we get $$\begin{align}x\cdot e_t&=\sum_{h\in C_t}\,\sum_{\tau\in T\cap R_{h\cdot t}}\,\text{sign}(h)\,\big(\tau\cdot \Omega_{h\cdot t}\big)-\sum_{h\in C_t}\,\sum_{\tau \in T\cap C_t}\,\text{sign}(\tau h)\,\big((\tau h)\cdot \Omega_t\big) \\&=\sum_{h\in C_t}\,\sum_{\tau\in T\cap R_{h\cdot t}}\,\text{sign}(h)\,\Omega_{h\cdot t}-\sum_{\tau \in T\cap C_t}\,\sum_{h\in C_t}\,\text{sign}(\tau h)\,\big((\tau h)\cdot \Omega_t\big) \\&=\sum_{h\in C_t}\,\text{sign}(h)\,\left|T\cap R_{h\cdot t}\right|\,\Omega_{h\cdot t}-\sum_{\tau \in T\cap C_t}\,\sum_{h\in C_t}\,\text{sign}(h)\,\left(h\cdot \Omega_t\right) \\&=\sum_{h\in C_t}\,\text{sign}(h)\,\left|T\cap R_{h\cdot t}\right|\left(h\cdot \Omega_t\right)-\left|T\cap C_t\right|\,\sum_{h\in C_t}\,\text{sign}(h)\,\left(h\cdot \Omega_t\right)\,. \end{align}$$
The rest is simple combinatorics. Note that $$\left|T\cap R_{h\cdot t}\right|=\sum_{k=1}^l\,\binom{\lambda_k}{2}\text{ for all }h\in C_t$$ and $$\left|T\cap C_t\right|=\sum_{k=1}^l\,(k-1)\,\lambda_k\,.$$ Consequently, $$x\cdot e_t=\left(\sum_{k=1}^l\,\binom{\lambda_k}{2}-\sum_{k=1}^l\,(k-1)\,\lambda_k\right)\,\sum_{h\in C_t}\,\text{sign}(h)\,(h\cdot \Omega_t)=\left(\sum_{k=1}^l\,\sum_{r=1}^{\lambda_k}\,(r-k)\right)\,e_t\,.$$ Ergo, we indeed have $$c(\lambda)=\sum_{k=1}^l\,\sum_{r=1}^{\lambda_k}\,(r-k)\,.$$
If I write $T^\mu$ for the sert of all permutations in $S_n$ of type $\mu\vdash n$, then $x^\mu:=\sum\limits_{s\in T^\mu}\,s$ is also a central element. It would be an interesting problem (probably known) to determine the action of $x^\mu$ on $V_\lambda$ for each $\lambda \vdash n$, which is multiplication by some scalar $c^\mu(\lambda)$. The center $\mathfrak{Z}\big(\mathbb{C}[S_n]\big)$ of $\mathbb{C}[S_n]$ is generated $x^{\mu(r)}$ where $\mu(r)=(r,1,1,\ldots,1)\vdash n$ for $r=1,2,\ldots,n$, if I remember correctly. In particular, if $n\geq 2$ and $\mu=\mu(2)$, then we already know that $$c^{\mu(2)}(\lambda)=\sum_{k=1}^l\,\binom{\lambda_k}{2}-\sum_{m=1}^p\,\binom{\lambda^\top_m}{2}\,,$$ where $\lambda^\top=\left(\lambda^\top_1,\lambda_2^\top,\ldots,\lambda^\top_p\right)$ is the transpose of $\lambda$. If $n\geq 3$ and $\mu=\mu(3)$, then I believe that $$c^{\mu(3)}(\lambda)=\sum_{k=1}^l\,\binom{\lambda_k}{3}+\sum_{m=1}^{p}\,\binom{\lambda^\top_m}{3}-\sum_{k=1}^l\,\sum_{m=1}^{p}\,(\lambda_k-1)(\lambda^\top_m-1)\,.$$ The proof is similar to $c^{\mu(2)}$, if I did not make any mistake.