The reason your argument doesn't work is because it's not true that $\text{Sym}^l \mathfrak n^\vee$ has a filtration with 1-dimensional graded pieces where $T$ acts with anti-dominant weights. In fact, this is false even in the case $l = 1$. I'll assume, as you do, that $B$ corresponds to the positive roots. Then the weights of $\mathfrak n^\vee$ correspond to the negative roots, and most of the negative roots are NOT anti-dominant. (In fact, even the simple roots are not antidominant, unless all components of $G$ are of type $A_1$). This shows why something subtle is going on here: $H^i( G/B, \mathfrak n^\vee ) = 0$ for all $i > 0$, but it is not the case in general that $H^i( G/B, \; \textrm{gr} \; \mathfrak n^\vee ) = 0$ for all $i > 0$.

The reason your argument doesn't work is because it's not true that $\text{Sym}^l \mathfrak n^\vee$ has a filtration with 1-dimensional graded pieces where $T$ acts with anti-dominant weights. In fact, this is false even in the case $l = 1$. I'll assume, as you do, that $B$ corresponds to the positive roots. Then the weights of $\mathfrak n^\vee$ correspond to the negative roots, and most of the negative roots are NOT anti-dominant. (In fact, even the negative simple roots are not antidominant, unless all components of $G$ are of type $A_1$). This shows why something subtle is going on here: $H^i( G/B, \mathfrak n^\vee ) = 0$ for all $i > 0$, but it is not the case in general that $H^i( G/B, \; \textrm{gr} \; \mathfrak n^\vee ) = 0$ for all $i > 0$.