After Phil's answer and the ensuing discussion, the remaining question can be formulated as:

Estimate $\lambda$ s.t. if $F\subset E$ and $E/F$ is finite dimensional, then every norm one linear operator $T$ from $F$ into a $C(K)$ space can be extended to an operator from $E$ into the $C(K)$ space which has norm at most $\lambda$.

I want to point out that $\lambda$ cannot be less than two.  I think that it is known that $\lambda$ can be anything larger than two, but I did not find a reference after a quick search.  I called Morry Zippin's attention to the problem; maybe he will comment. Or maybe this was proved after he worked on almost complementation; perhaps by Jesus Castillo and colleagues.  I more or less thought through that $\lambda$ can be anything larger than three. 

Before saying why $\lambda$ cannot be less than two, let's further reformulate the problem.  The second dual of every $C(K)$ is $1$-injective, so the problem reduces to the case where $F$ is a $C(K)$ space and $T$ is the identity operator on $F$. That is, we need to estimate how well complemented a $C(K)$ must be in a superspace in which the $C(K)$ space has finite codimension. In fact, the superspace of $C(K)$ can be assumed to be in $C(K)^{**}$, but that is not particularly useful in getting the lower bound on $\lambda$. 

Consider the $C(K)$ space $c$, the space of convergent sequences under the sup norm. (I'll treat the real case, but the complex case is only slightly more complicated.)  The superspace $E$ is the span in  $\ell_\infty$ of $c$ and the sequence $x$ defined by $x(n)= (-1)^n$.  Let $P_n$ be the projection onto the span of the first $n$ unit vectors in $\ell_\infty$ and let $P$ be any projection from $E$ onto $c$.  Write $Px = a\cdot 1 + y$ with $y$ in $c_0$ and where $1$ is the constant one function. Now for every $n$, the norm of $x-2P_n x$ is one, and $P(x- 2P_n x)=a 1 + y - 2P_n x$.  Since $y$ is in $c_0$, letting $n\to \infty$ gives $\|P\| \ge |a|+2$.  The choice $a=0$, $y=0$ of course gives a projection of norm two.