From JS1's answer we already see that this is impossible for normal dice. But assuming the dice's result are dependent on each other as if they could communicate,
there is indeed a way to realize this.
This is how it works:
Since the probabilities are dependent on the outcome of the other dice $P(\color{red}{i}, \color{green}{j}) \neq P(\color{red}{i}) \cdot P(\color{green}{j})$. This means we have all 36 $P(\color{red}{i}, \color{green}{j})$ as free variables and only 11 conditions. Ok, admittedly, we must make sure that all probabilities are within $\left[ 0, 1 \right]$, but this is just restricting the hypervolume of the 25-dimensional solution space and not reducing its dimensionality.
So there are infinitely many possibilities for solutions.
Let's pick a "nice" solution:
As additional criterion I want that
- the dice are equal: $P(\color{red}{i}, \color{green}{j}) = P(\color{red}{j}, \color{green}{i})$
- and they should behave like normal dice, i.e. it shows each side with equal probability if you ignore the outcome of the other dice: $\color{green}{\sum_{j = 1}^6} P(\color{red}{i}, \color{green}{j}) = \frac{1}{6}$
Like this they appear as if they are normal dice. Only if you look at the correlations you find that something is weird.
Actual numbers:
All possibilities which lead to a sum of $\color{#26F}{2}$ are $\left\{ (\color{red}{1}, \color{green}{1}) \right\}$, so we know immediately that the corresponding probability must be $\frac{1}{11}$. The same is true for a sum of $\color{#26F}{12}$, because here we also have only $1$ possible way $\left\{ (\color{red}{6}, \color{green}{6}) \right\}$.
$$P(\color{red}{1}, \color{green}{1}) = \frac{1}{11} = P(\color{red}{6}, \color{green}{6})$$
For a sum of $\color{#26F}{3}$ there are $2$ possibilities: $\left\{ (\color{red}{1}, \color{green}{2}), (\color{red}{2}, \color{green}{1}) \right\}$, but since the dice are equal they appear with equal probability. Furthermore they must add up to $\frac{1}{11}$. The same holds for a sum of $\color{#26F}{11}$.
$$P(\color{red}{1}, \color{green}{2}) = P(\color{red}{2}, \color{green}{1}) = \frac{1}{11} \cdot \frac{1}{2} = P(\color{red}{5}, \color{green}{6}) = P(\color{red}{6}, \color{green}{5})$$
$3$ different outcomes lead to sums of $\color{#26F}{4}$ and $\color{#26F}{10}$, respectively. The probabilities for $(\color{red}{1}, \color{green}{3})$ and $(\color{red}{3}, \color{green}{1})$ are equal, but the probability for $(\color{red}{2}, \color{green}{2})$ can be different from the other two. Let's parametrize this degree of freedom by $a$.
$$P(\color{red}{1}, \color{green}{3}) = P(\color{red}{3}, \color{green}{1}) = \frac{1}{11} \cdot a = P(\color{red}{4}, \color{green}{6}) = P(\color{red}{6}, \color{green}{4}) \\ P(\color{red}{2}, \color{green}{2}) = \frac{1}{11} \cdot (1-2a) = P(\color{red}{5}, \color{green}{5})$$
It continues similarly for sums of $\color{#26F}{5}$ and $\color{#26F}{9}$, except that here we have $4$ possibilities each, with pairs of them being equal.
$$P(\color{red}{1}, \color{green}{4}) = P(\color{red}{4}, \color{green}{1}) = \frac{1}{11} \cdot b = P(\color{red}{3}, \color{green}{6}) = P(\color{red}{6}, \color{green}{3}) \\ P(\color{red}{2}, \color{green}{3}) = P(\color{red}{3}, \color{green}{2}) = \frac{1}{11} \cdot \frac{1}{2} (1-2b) = P(\color{red}{4}, \color{green}{5}) = P(\color{red}{5}, \color{green}{4})$$
For sums of $\color{#26F}{6}$ and $\color{#26F}{8}$ there are $5$ possibilities each, hence $3$ independent numbers, which means $2$ more parameters.
$$P(\color{red}{1}, \color{green}{5}) = P(\color{red}{5}, \color{green}{1}) = \frac{1}{11} \cdot c = P(\color{red}{2}, \color{green}{6}) = P(\color{red}{6}, \color{green}{2}) \\ P(\color{red}{2}, \color{green}{4}) = P(\color{red}{4}, \color{green}{2}) = \frac{1}{11} \cdot d = P(\color{red}{3}, \color{green}{5}) = P(\color{red}{5}, \color{green}{3}) \\ P(\color{red}{3}, \color{green}{3}) = \frac{1}{11} \cdot (1 - 2c - 2d) = P(\color{red}{4}, \color{green}{4})$$
Finally, for a sum of $\color{#26F}{7}$ there are $6$ possibilities, hence another $2$ parameters.
$$P(\color{red}{1}, \color{green}{6}) = P(\color{red}{6}, \color{green}{1}) = \frac{1}{11} \cdot e \hphantom{= P(\color{red}{1}, \color{green}{6}) = P(\color{red}{6}, \color{green}{1})} \\ P(\color{red}{2}, \color{green}{5}) = P(\color{red}{5}, \color{green}{2}) = \frac{1}{11} \cdot f \hphantom{= P(\color{red}{2}, \color{green}{5}) = P(\color{red}{5}, \color{green}{2})} \\ P(\color{red}{3}, \color{green}{4}) = P(\color{red}{4}, \color{green}{3}) = \frac{1}{11} \cdot \frac{1}{2} (1 - 2c - 2d) \hphantom{= P(\color{red}{4}, \color{green}{4})}$$
Now let's implement the additional condition 2.
$$\begin{align} \frac{1}{6} &= \color{green}{\sum_{j=1}^6} P(\color{red}{1}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{1}) = \color{green}{\sum_{j=1}^6} P(\color{red}{6}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{6}) \\ &= \frac{1}{11} \left( 1 + \frac{1}{2} + a + b + c + e \right) \\ \frac{1}{6} &= \color{green}{\sum_{j=1}^6} P(\color{red}{2}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{2}) = \color{green}{\sum_{j=1}^6} P(\color{red}{5}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{5}) \\ &= \frac{1}{11} \left( \frac{1}{2} + (1 - 2a) + \frac{1}{2} (1 - 2b) + c + d + f \right) \\ \frac{1}{6} &= \color{green}{\sum_{j=1}^6} P(\color{red}{3}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{3}) = \color{green}{\sum_{j=1}^6} P(\color{red}{4}, \color{green}{j}) = \color{red}{\sum_{i=1}^6} P(\color{red}{i}, \color{green}{4}) \\ &= \frac{1}{11} \left( a + \frac{1}{2} (1 - 2b) + (1 - 2c - 2d) + \frac{1}{2} (1 - 2e - 2f) + c + d \right) \end{align}$$
This can be further reduced to
$$\begin{align} e &= \frac{1}{3} - a - b - c \\ f &= 2a + b - c - d - \frac{1}{6} \text{.} \end{align}$$
By fixing 4 of these parameters we can find a concrete solution. For example $a = \frac{1}{3}$, $b = c = d = e = 0$, $f = \frac{1}{2}$. This gives the following probabilities: