First, by a self-similarity argument, it suffices to consider the case $T=1$. We can use the equivalence of the usual Hölder norm with the sequence norm, defined by $$ \lVert x\rVert_\alpha:=\sup_{j\geqslant 1}2^{j\alpha}\max_{1\leqslant k\leqslant 2^{j-1}}\left\lvert x\left(\left(2k\right)2^{-j}\right)-2x\left(\left(2k-1\right)2^{-j}\right) +x\left(\left(2k-2\right)2^{-j}\right)\right\rvert +\left\lvert x\left(0\right)\right\rvert+\left\lvert x\left(1\right)\right\rvert.$$ This was established in Ciesielski, Z. (1960). On the isomorphisms of the spaces Hα and m. Bull. Acad. Pol. Sci. Ser. Sci. Math. Phys. 8, 217–222.

Applying this to our problem, it suffices to consider the moments of the random variable $$ Y:=\sup_{j\geqslant 1}2^{j\alpha}\max_{1\leqslant k\leqslant 2^{j-1}}\left\lvert B\left(\left(2k\right)2^{-j}\right)-2B\left(\left(2k-1\right)2^{-j}\right) +B\left(\left(2k-2\right)2^{-j}\right)\right\rvert. $$ We control for a fixed $t$ the probability that $Y\gt 2t$ in the following way: $$ \Pr\left(Y\gt 2t\right)\leqslant \sum_{j=1}^{+\infty}\sum_{k=1}^{2^{j-1}} \Pr\left(\left\lvert B\left(\left(2k\right)2^{-j}\right)-2B\left(\left(2k-1\right)2^{-j}\right) +B\left(\left(2k-2\right)2^{-j}\right)\right\rvert\gt 2t2^{-j\alpha}\right). $$ Using stationarity of increments of standard Brownian motion, we derive the bound $$ \Pr\left(Y\gt 2t\right)\leqslant \sum_{j=1}^{+\infty} 2^{j } \Pr\left(\left\lvert B\left( 2^{-j}\right) \right\rvert\gt t2^{-j\alpha}\right). $$ By self-similarity, we derive that $$ \Pr\left(Y\gt 2t\right)\leqslant \sum_{j=1}^{+\infty} 2^{j } \Pr\left(\left\lvert N\right\rvert\gt t2^{j\left(1/2-\alpha\right)}\right), $$ where $N$ has a standard normal distribution. A comparison between series and integrals yields $$ \Pr\left(Y\gt 2t\right)\leqslant \mathbb E\left[ \left\lvert N\right\rvert^{\frac 1{1/2-\alpha}}x^{-\frac 1{1/2-\alpha}} \mathbb 1\left\{\left\lvert N\right\rvert \gt t\right\} \right]. $$ Multiplying by $t^{p-1}$ and integrating over $(1,+\infty)$ shows that the Hölderian norm has moments of all orders. Probably more can be deduced from the point of view of exponential moments.

First, by a self-similarity argument, it suffices to consider the case $T=1$. We can use the equivalence of the usual Hölder norm with the sequence norm, defined by $$ \lVert x\rVert_\alpha:=\sup_{j\geqslant 1}2^{j\alpha}\max_{1\leqslant k\leqslant 2^{j-1}}\left\lvert x\left(\left(2k\right)2^{-j}\right)-2x\left(\left(2k-1\right)2^{-j}\right) +x\left(\left(2k-2\right)2^{-j}\right)\right\rvert +\left\lvert x\left(0\right)\right\rvert+\left\lvert x\left(1\right)\right\rvert.$$ This was established in Ciesielski, Z. (1960). On the isomorphisms of the spaces Hα and m. Bull. Acad. Pol. Sci. Ser. Sci. Math. Phys. 8, 217–222.

Applying this to our problem, it suffices to consider the moments of the random variable $$ Y:=\sup_{j\geqslant 1}2^{j\alpha}\max_{1\leqslant k\leqslant 2^{j-1}}\left\lvert B\left(\left(2k\right)2^{-j}\right)-2B\left(\left(2k-1\right)2^{-j}\right) +B\left(\left(2k-2\right)2^{-j}\right)\right\rvert. $$ We control for a fixed $t$ the probability that $Y\gt 2t$ in the following way: $$ \Pr\left(Y\gt 2t\right)\leqslant \sum_{j=1}^{+\infty}\sum_{k=1}^{2^{j-1}} \Pr\left(\left\lvert B\left(\left(2k\right)2^{-j}\right)-2B\left(\left(2k-1\right)2^{-j}\right) +B\left(\left(2k-2\right)2^{-j}\right)\right\rvert\gt 2t2^{-j\alpha}\right). $$ Using stationarity of increments of standard Brownian motion, we derive the bound $$ \Pr\left(Y\gt 2t\right)\leqslant \sum_{j=1}^{+\infty} 2^{j } \Pr\left(\left\lvert B\left( 2^{-j}\right) \right\rvert\gt t2^{-j\alpha}\right). $$ By self-similarity, we derive that $$ \Pr\left(Y\gt 2t\right)\leqslant \sum_{j=1}^{+\infty} 2^{j } \Pr\left(\left\lvert N\right\rvert\gt t2^{j\left(1/2-\alpha\right)}\right), $$ where $N$ has a standard normal distribution. A comparison between series and integrals yields $$\tag{*} \Pr\left(Y\gt 2t\right)\leqslant \mathbb E\left[ \left\lvert N\right\rvert^{\frac 1{1/2-\alpha}}t^{-\frac 1{1/2-\alpha}} \mathbb 1\left\{\left\lvert N\right\rvert \gt t\right\} \right]. $$ Multiplying by $t^{p-1}$ and integrating over $(1,+\infty)$ shows that the Hölderian norm has moments of all orders. More can be deduced from the point of view of exponential moments. Indeed, for $\varepsilon\lt 1/2$, $$ \mathbb{E} \left[\exp\left(\varepsilon \|B\|_{\alpha}^2\right)\right] < \infty. $$ Multypling indeed in (*) by the derivative of $t\mapsto \exp\left(\varepsilon t^{2}\right)$ and integrate on $(1,+\infty)$.