The estimate is too generous! The idea, behind the problem is an well known extremal property of the Chebyshev polynomials. Among all polynomials of degree n and with magnitude at most $1$ in $[ -1,1]$, the Chebyshev one has the greatest magnitude outside [-1,1]. More precisely Let $ P(x)$ be a polynomial of degree $ n$ with $ |P(x)|\leq 1, \forall x\in[-1,1]$ and $ T_n(x)$ be the Chebyshev polynomial of degree $ n$. Then $$ |P(x)|\leq |T_n(x)|, \forall x\notin [-1,1]$$This claim has a surprisingly simple proof, based on the fact the Chebyshev polynomial oscillates neatly in $[-1,1]$. In our case, all we need is to transform $P$ on $[-1,1]$ and apply it. I had written something about this extremal property in my blog (theorem 2 can be applied directly for the OP).

By the Lagrange Interpolation formula, $f\left(-\frac{1}{n}\right) = \sum_{k=0}^{n} f\left(\frac{k}{n}\right) \: \prod_{j \neq k}^{n} \frac{-(j+1)}{k-j}$ and then, $|f\left(-\frac{1}{n}\right)| \le \sum_{k=0}^{n} f\left(\frac{k}{n}\right) \: \prod_{j \neq k}^{n} \frac{(j+1)}{|k-j|} =\sum_{k=0}^{n} {n+1\choose k+1} = 2^{n+1}-1$ by the triangle inequality $\blacksquare$