Variance of a random variable representing the sum of two dice $\newcommand{\Var}{\operatorname{Var}}$

The formula for the variance of the sum of two independent random variables is given $$ \Var (X +X) = \Var(2X) = 2^2\Var(X)$$

How then, does this happen:

Rolling one dice, results in a variance of $\frac{35}{12}$. Rolling two dice, should give a variance of $2^2\Var(\text{one die}) = 4 \times \frac{35}{12} \approx 11.67$. Instead, my Excel spreadsheet sample (and other sources) are giving me 5.83, which can be seen is equal to only $2 \times \Var(X)$. 

What am I doing wrong?
$\newcommand{\Var}{\operatorname{Var}}$

The formula you give is not for two independent random variables.  It's for random variables that are as far from independent as you can get.  If $X,Y$ are independent, then you have $\Var(X+Y)=\Var(X)+\Var(Y)$.  If, in addition, $X$ and $Y$ both have the same distribution, then this is equal to $2\Var(X)$.  It is also the case that, as you say, $\Var(X+X)=4\Var(X)$.  But that involves random variables that are nowhere near independent.
X and itself aren't independent. X and itself are correlated with rho = 1. 

In general, for two events, X (the first die) and Y (the second die), VAR(X+Y) = VAR(X) + VAR(Y) + 2COV(XY). 

Since rolling two dice ARE independent (rho = 0), COV(XY) = 0, so VAR(X+Y) = 35/12 + 35/12 + 0. 

VAR(X+X) is 4*VAR(X), but this recognizes that X and itself are perfectly correlated and we could just as well calculate this as VAR(X+X) = 35/12 + 35/12 + 2*35/12 since we know that COV(XX) is the same as VAR(X).