We may assume that $l\not\equiv 0\pmod p$ because otherwise the given sum is simple. The answer is a [Kloosterman sum][1].

Let $$\delta_q(x)=\begin{cases}
1,& \text{if } x\equiv 0\pmod{q};\\
0,& \text{if } x\not\equiv 0\pmod{q}.\\
\end{cases}$$
Then
\begin{gather*}
S(l)=\sum\limits_{k=1}^{p}\left(\dfrac{k(k+1)}{p}\right)e\left(\dfrac{kl}{p}\right)=\\=
\sum\limits_{k=1}^{p}\left(\sum\limits_{y=1}^{p}\delta_p(k(k+1)-y^2)-1\right)e\left(\dfrac{kl}{p}\right)=\\=
\sum\limits_{k,y=1}^{p}\delta_p(k(k+1)-y^2)e\left(\dfrac{kl}{p}\right)=[k=x+y]=\\=
\sum\limits_{x,y=1}^{p}\delta_p(x^2+2xy+x+y)e\left(\dfrac{(x+y)l}{p}\right).
\end{gather*}
For each non-zero summand $y=\dfrac{x^2+x}{2x+1}$. Hence
\begin{gather*}
S(l)=
\sum\limits_{x=1}^{p}e\left(\dfrac{l}{p}\cdot\left(\dfrac{x^2+x}{2x+1}+x\right)\right)=[t=2x+1]=\\=
\sum\limits_{t=1}^{p}e\left(\dfrac{l}{p}\cdot\left(\dfrac{3t}{4}-\dfrac{1}{4t}-\dfrac{1}{2}\right)\right).
\end{gather*}


  [1]: http://en.wikipedia.org/wiki/Kloosterman_sum