First a general comment: as Sasha alludes to, there are two tensor products of complexes of sheaves. Let $i : X \hookrightarrow X \times X$ denote the diagonal. We have, and given complexes of sheaves $\mathcal{F}$ and $\mathcal{G}$ on $X$ we can form their external tensor product $\mathcal{F} \boxtimes \mathcal{G}$ on $X \times X$. The two tensor products are
\begin{gather*}
\mathcal{F} \otimes \mathcal{G} := i^*(\mathcal{F} \boxtimes \mathcal{G}) \quad \text{and} \\
\mathcal{F} \otimes^{!} \mathcal{G} := i^!(\mathcal{F} \boxtimes \mathcal{G}).
\end{gather*}
(I think I learnt this notation from some very nice old notes of Ginzburg.) This description makes it clear that $\mathbb{D}(\mathcal{F} \otimes \mathcal{G}) = \mathbb{D}\mathcal{F} \otimes^{!} \mathbb{D}\mathcal{G}$ etc. In particular, we have control of the stalks of $\mathcal{F} \otimes \mathcal{G}$ in terms of the stalks of $\mathcal{F}$ and $\mathcal{G}$, and costalks of $\mathcal{F} \otimes^{!} \mathcal{G}$ in terms of costalks of $\mathcal{F}$ and $\mathcal{G}$, but for example the costalks of $\mathcal{F} \otimes \mathcal{G}$ are potentially very complicated.

Now to your specific question: consider the big tilting sheaf $T_s$ on $\mathbb{P}^1$. (This is isomorphic to the projective cover you ask about.) Its stalks are $\mathbb{Q}[1]$ on the open locus, and $\mathbb{Q}$ at the point stratum. In particular, the stalks of $T_s \otimes T_s$ are $\mathbb{Q}[1] \otimes \mathbb{Q}[1] = \mathbb{Q}[2]$ on the open locus, and $\mathbb{Q}$ on the point stratum. In particular we have a distinguished triangle
\begin{equation}
j_!\mathbb{Q}[2] \to T_s^{\otimes 2} \to \mathbb{Q}_{0} \stackrel{[1]}{\to}
\end{equation}
where $j$ denotes the inclusion of the open stratum, and $0$ denotes the point stratum. Because the appropriate Ext group vanishes, we deduce that
\begin{equation}
T_s^{\otimes 2} \cong j_!\mathbb{Q}[2] \oplus \mathbb{Q}_{0}.
\end{equation}
(Already in this example we see the failure of the tensor product to be self-dual.)

What about in general? We know the stalks of tilting sheaves on $G/B$ (see Yun's "weights of mixed tilting sheaves" for a lovely account). They are concentrated in a single degree, with dimension given by Kazhdan-Lusztig polynomials at 1. Hence the stalks of $T_x \otimes T_y$ along the stratum $BzB/B$ are concentrated in degree $-2\ell(z)$ and are computable via a product of KL polynomials at $1$. I am not sure if the tensor product splits as above. But maybe that is your job :)

An aside: if one instead considers tensor products of $IC$ sheaves several small example suggest that the answer is always pure. (This has gotten me excited at least twice in my life, and I know I am not the only one!) However this is not the case in general, as the example of $IC_{st} \otimes IC_{ts}$ in $SL_3/B$ shows. (In this example the tensor product is the constant sheaf on the union of the two closed Schubert curves corresponding to $s$ and $t$.)