Thanks to the hints of Narutaka Ozawa and Yemon Choi I can answer this question by myself.

Recall that $I^{\perp\perp}$ is the weak${}^*$ closure of $I$ in $A^{**}$, so by Goldstine theorem we can choose a net $(e_\nu'')_{\nu\in N''}\subset I$ such that weeak${}^*$ converges to $e$. Clearly $(1-e_\nu'')_{\nu\in N''}$ converges to $1-e$ in the same topology. By lemma 1.1 from [this paper][1] there exists a net in $\operatorname{conv}(1-e_\nu'')_{\nu\in N''}=1-\operatorname{conv}(e_\nu'')_{\nu\in N''}$ that weak${}^*$ converges to $1-e$ with norm bound $\Vert 1-e\Vert$. Denote this net as $(1-e_\nu')_{\nu\in N'}$, then it is easy to check that $(e_\nu')_{\nu\in N'}$ weak${}^*$ converges to $e$ and a weak right approximate identity for $I$. By proposition 33.2 in Approximate identities and factorization in Banach algebras by Doran R. S., Wichmann J. there is a net $(e_\nu)_{\nu\in N}\subset\operatorname{conv}(e_\nu')_{\nu\in N'}$ which is a right bounded approximate identity for $I$. For any $\nu\in N$ the vector $1-e_\nu\in\operatorname{conv}(1-e_\nu')_{\nu\in N'}$, then taking into account the norm bound on $(1-e_\nu')_{\nu\in N'}$ we get 
$$
\sup_{\nu\in N}\Vert 1-e_\nu\Vert
\leq\Vert 1-e\Vert
$$


  [1]: http://arxiv.org/abs/1405.5551