**Not a full proof**, but a number of thoughts too long for a comment. This post aims at finding alternative (yet harder) criteria for proving the conjecture. Please discuss.

As in previous comments, let's denote
$
X=6(A^3+B^3+C^3)+I
$
and
$
Y=5(A^2+B^2+C^2)
$ and $D = X-Y$.

The question is to show $\det(X) \ge \det(Y)$ or $1 \ge \det(X^{-1}Y)$. Write $Q = X^{-1}Y$, then $Q$ is positive definite, since $X$ and $Y$ are positive definite. Now it is known for a positive definite matrix $Q$ (see e.g. here) that the *trace bound* is given by
$$
\bigg(\frac{\text{Tr}(Q)}{n}\bigg)^n \geq \det(Q)
$$

So a second (harder) criterion for the conjecture is $n \ge \text{Tr}(X^{-1}Y)$ or $ \text{Tr}(X^{-1}D) \geq 0$. I wouldn't see how I can compute this trace or find bounds, can someone?

Let's call $d_i$ the eigenvalues of $D$, likewise for $X$ and $Y$. While $x_i > 0$, this doesn't necessarily hold for $d_i$ since we know from comments that $D$ is not necessarily positive definite. So (if $X$ and $D$ could be simultaneously diagonalized) $ \text{Tr}(X^{-1}D) = \sum_i \frac{d_i}{x_i} = r \sum_i {d_i}$ where there exists an $r$ by the mean value theorem. Where $r$ is not guaranteed to be positive, it is likely that $r$ *will* be positive, since $r$ will only become negative if there are (many, very) negative $d_i$ with small associated $x_i$. Can positivity of $r$ be shown? If we can establish that a positive $r$ can be found,
a third criterion is $ \text{Tr}(D) \geq 0$.

Now with this third criterion, we can use that the trace is additive and that the trace of commutors vanishes, i.e. $\text{Tr} (AB -BA) = 0$. Using this argument, it becomes unharmful when matrices do not commute, since under the trace their order can be changed. This restores previous solutions where the conjecture was reduced to the valid Schur's inequality (as noted by a previous commenter), which proves the conjecture.

A word on how hard the criteria are, indicatively in terms of eigenvalues:

(hardest) positive definiteness: $d_i >0$ $\forall i$ or equivalently, $\frac{y_i}{x_i} <1$ $\forall i$

(second- relies on positive $r$) $ \text{Tr}(D) \geq 0$: $\sum_i d_i \geq 0$

(third) $n \ge \text{Tr}(X^{-1}Y)$: $\sum_i \frac{y_i}{x_i} \leq n$

(fourth - least hard) $\det(X) \ge \det(Y)$: $\prod_i \frac{y_i}{x_i} \leq 1$

Solutions may also be found by using criteria which interlace between those four.

A word on simulations and non-positive-definiteness:

I checked the above criteria for the non-positive definite example given by @user1551 in the comments above, and the second, third and fourth criteria hold.

Note that equality $\det(X) = \det(Y)$ occurs for (a) symmetry point: $A=B=C=\frac13 I$ and for (b) border point: $A=B=\frac12 I$ and $C=0$ (and permutations). I checked the "vicinity" of these equality points by computer simulations for real matrices with $n=2$ where I extensively added small matrices with any parameter choices to $A$ and $B$ (and let $C = I - A-B$), making sure that $A,B$ and $C$ are positive definite. It shows that for the vicinity of the symmetry point, the second, third and fourth criteria above hold, while there occur frequent non-positive-definite examples. For the vicinity of the border point all four criteria hold.